Question 13 Marks
An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
AnswerHeat is supplied to the system at a rate of 100W. $\therefore$ Heat supplied, Q = 100J/s The system performs at a rate of 75J/s. $\therefore$ Work done, W = 75J/s From the first law of thermodynamics, we have: Q = U + W Where, U = Internal energy$\therefore$ U = Q - W
= 100 - 75 = 25J/s = 25W Therefore, the internal energy of the given electric heater increases at a rate of 25W.
View full question & answer→Question 23 Marks
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig.
Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.
AnswerTotal work done by the gas from D to E to $\mathrm{F}=$ Area of $\triangle \mathrm{DEF}$ Area of $\triangle \mathrm{DEF}=(1 / 2) \mathrm{DE} \times \mathrm{EF}$ Where, $\mathrm{DF}=$ Change in pressure $=600 \mathrm{~N} / \mathrm{m}^2-300 \mathrm{~N} / \mathrm{m}^2=300 \mathrm{~N} / \mathrm{m}^2 \mathrm{FE}=$ Change in volume $=5.0 \mathrm{~m}^3-2.0 \mathrm{~m}^3=3.0 \mathrm{~m}^3$ Area of $\triangle \mathrm{DEF}=$ $(1 / 2) \times 300 \times 3=450$ J Therefore, the total work done by the gas from $D$ to $E$ to $F$ is 450 J.
View full question & answer→Question 33 Marks
Explain why: Two bodies at different temperatures $T_1$ and $T_2$ if brought in thermal contact do not necessarily settle to the mean temperature $(T_1 + T_2 )/2$.
AnswerWhen two bodies at different temperatures $T_1$ and $T_2$ are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature $(T_1 + T_2)/2$ only when the thermal capacities of both the bodies are equal.
View full question & answer→Question 43 Marks
A refrigerator is to maintain eatables kept inside at $9^{\circ} \mathrm{C}$. If room temperature is $36^{\circ} \mathrm{C}$, calculate the coefficient of performance.
AnswerTemperature inside the refrigerator, $\mathrm{T}_1=9^{\circ} \mathrm{C}=282 \mathrm{~K}$ Room temperature, $\mathrm{T}_2=36^{\circ} \mathrm{C}=309 \mathrm{~K}$ Coefficient of performance $=T_1 / T_2-T_1=282 / 309-282=282 / 27=10.44$ Therefore, the coefficient of performance of the given refrigerator is 10.44.
View full question & answer→Question 53 Marks
A steam engine delivers $5.4 \times 10^8J$ of work per minute and services $3.6 \times 10^9J$ of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
AnswerWork done by the steam engine per minute, $\mathrm{W}=5.4 \times 10^8 \mathrm{~J}$ Heat supplied from the boiler, $\mathrm{H}=3.6 \times 10^9 \mathrm{~J}$ Efficiency of the engine = output energy/Input energy.
$\therefore \mathrm{n}=\mathrm{W} / \mathrm{H}$
$=5.4 \times 10^8 / 3.6 \times 10^9$
Hence, the percentage efficiency of the engine is $15 \%$. Amount of heat wasted $=3.6 \times 10^9$
$5.4 \times 10^8=30.6 \times 10^8=3.06 \times 10^9 \mathrm{~J}$
Therefore, the amount of heat wasted per minute is $3.06 \times 10^9 \mathrm{~J}$.
View full question & answer→Question 63 Marks
State first law of thermodynamics. What are its limitations? Why $C_p > C_v$?
AnswerAccording to first law of thermodynamics, the total heat energy change $dQ$ is the sum of internal energy change $dU$ and work done $dW$. i.e. $dQ = dU + DW$
Limitations:
First law do not tell us,
- The quick or slow nature of a process.
- Whether the process is possible or not. $C_p > C_v$ because for constant pressure process, both volume and temperature are altered and for constant volume process only temperature varies.
View full question & answer→Question 73 Marks
The volume of an ideal gas is V at a pressure P. On increasing the pressure by $\Delta\text{P},$ the change in volume of the gas is $(\Delta\text{V}_1)$ under isothermal conditions and $(\Delta\text{V}_2)$ under adiabatic conditions. Is $\Delta\text{V}_1>\Delta\text{V}_2$ or vice-versa and why?
AnswerUnder isothermal condition, $\text{K}_\text{i}=\frac{\Delta\text{P}}{\frac{\Delta\text{V}_1}{\text{V}}}=\text{p}\ ...(\text{i)}$ Under adiabatic condition, $\text{K}_\text{a}=\frac{\Delta\text{P}}{\Delta\text{V}_2\text{V}}=\gamma\text{P}\ ...(2)$ Dividing (ii) by (i), we get , $\frac{\Delta\text{}V_1}{\Delta\text{V}_2}\gamma$ As $\gamma>1,$ $(\Delta\text{V}_1)>(\Delta\text{V}_2).$
View full question & answer→Question 83 Marks
Calculate the fall in temperature when a gas initially at $72°C$ is expanded suddenly to eight times its original volume. Given y $=\frac53.(\therefore\text{V}_2=8\text{x}\text{ c.c.})$
AnswerLet, $V_1 = x c.c.; T_1 = 273 + 72 = 345K; \gamma=\frac53; T_2 =$ ? Using the relation $\text{T}_1\text{V}^{\gamma-1}_1=\text{T}_2\text{V}^{\gamma-1}_2$
$\therefore\text{T}_2=\text{T}_1\Big(\frac{\text{V}_1}{\text{V}_2}\Big)^{\gamma-1}$ $=345\times\Big(\frac{\text{x}}{8\text{x}}\Big)^\frac23$
$=345\times\Big(\frac18\Big)^\frac23$
Taking $\log$ both sides, we get $\log\text{T}_2=\log\ 345-\frac23\log8$
$=2.5378-\frac23(0.9031)$
$=2.5378-0.6020$ $=1.9358$ $\text{T}_2=86.26\text{ K}$
$\therefore$ Fall in temperature = 345 - 86.26 = 258.74K.
View full question & answer→Question 93 Marks
Consider one mole of perfect gas in a cylinder of unit cross-section with a piston attached $($figure$)$. A spring $($spring constant $k)$ is attached $($unstretched length $L)$ to the piston and to the bottom of the cylinder. Initially, the spring is unstretched and the gas is in equilibrium. A certain amount of heat $Q$ is supplied to the gas causing an increase of volume from $V_0$ to $V_1$.
- What is the initial pressure of the system?
- What is the final pressure of the system?
- Using the first law of thermodynamics, write down a relation between $Q, P_a, V, V_0$, and $k$.
Answer
- $p_i = p_a$
- $\text{p}_\text{f}=\text{p}_\text{a}+\frac{\text{k}}{\text{A}}(\text{V}-\text{V}_0)$
- According to first law of thermodynamics,
$\Delta\text{Q}=\Delta\text{U}+\Delta\text{W}$ Where,
$\Delta\text{U}=\text{C}_\text{V}(\text{T}-\text{T}_0),$ $\Delta\text{W}=\text{p}_\text{a}(\text{V}-\text{V}_0)+\frac12\text{k}(\text{V}-\text{V}_0)^2$
$\Delta\text{Q}=\text{p}_\text{a}(\text{V}-\text{V}_0)+\frac12\text{k}(\text{V}-\text{V}_0)^2+\text{C}_\text{V}(\text{T}-\text{T}_0)$
Where, $\text{T}_0=\text{p}_\text{a}\frac{\text{V}_0}{\text{R}},$
$\Rightarrow\text{T}=\Big[\text{p}_\text{a}+\Big(\frac{\text{K}}{\text{A}}\Big)\times(\text{V}-\text{V}_0)\Big]\frac{\text{V}}{\text{R}}$ View full question & answer→Question 103 Marks
The initial state of a certain gas is $(P_i , V_i , T_i )$. It undergoes expansion till its volume becoms $V_f$. Consider the following two cases:
- The expansion takes place at constant temperature.
- The expansion takes place at constant pressure.
Plot the $P-V$ diagram for each case. In which of the two cases, is the work done by the gas more? 
Answer
- a. The expension from $V_i$ to Vf tempreature $T_i$ remains constant so isothermal expension i.e. $\mathrm{P}_{\mathrm{i}} \mathrm{V}_{\mathrm{i}}=\mathrm{P}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}$ constant $T$.
$b.$ The expension is at constant pressure $\mathrm{p}_{\mathrm{i}}$ so isobaric process so graph $P - V$ will be parallel to $V$ axis till its volume becomes $\mathrm{V}_{\mathrm{f}}$ As the area enclosed by graph $(a)$ is less than $(b)$ with volume axis so $W.D$. by process $(b)$ is more than of $(a).$
View full question & answer→Question 113 Marks
One mole of an ideal gas requires $207J$ heat to raise the temperature by $10K$ when heated at constant pressure. Find the amount of heat required to heat the same gas to raise the temperature by same $10K$ under constant volume conditions. Given $R = 8.3J mol^{-1} K^{-1}$.
AnswerHere heat required to raise temperature of 1 mole of gas through 10K under constant pressure conditions $\Delta\text{Q}=207\ \text{J}$
$\therefore\ \text{C}_\text{p}=\frac{\Delta\text{Q}}{\pi.\Delta\text{T}}$
$=\frac{207}{1\times10}$
$=20.7\ \text{J}\ \text{mol}^{-1}\text{K}^{-1}$
$\therefore\ \text{C}_\text{v}=\text{C}_\text{p}-\text{R}$
$=20.7-8.3$
$=12.4\ \text{J}\ \text{mol}^{-1}\text{K}^{-1}$
$\therefore$ Amount of heat required to raise the temperature of gas through 10K under constant volume condition: $\Delta\text{Q}'=\mu.\text{C}_\text{v}.\Delta\text{T}$
$=1\times12.4\times10=124\ \text{J}\ \text{mol}^{-1}\text{K}^{-1}$
View full question & answer→Question 123 Marks
A carnot engine with the cold body temperature of 17°C has 50% efficiency. By how much should the temperature of its hot body be changed to increase the efficiency to 60%? When will its efficiency be 100%?
Answer$\text{T}_2=273+17=290\text{K},\eta=50\%=\frac12$ As, $\eta=1-\frac{\text{T}_2}{\text{T}_1},\frac12=1-\frac{290}{\text{T}_1}$ $\text{T}_1=580\text{K}$ Again $\eta'=\text{60%}=\frac35$ Thus, $\eta'=1-\frac{\text{T}_2}{\text{T}_1},$ or $\frac35=1-\frac{290}{\text{T}_1'}$ $\text{T}_1'=725\text{K}$ Thus, $\text{T}_1'-\text{T}_1=725\text{K}-580\text{K}=145\text{K}$ For $\eta=1,1-\frac{\text{T}_2}{\text{T}_1}=1$ or $\text{T}_2=0\text{K}$
View full question & answer→Question 133 Marks
What are reversible and irreversible processes? Explain giving one example of each.
AnswerReversible Process: It is a process which can be made to proceed in two opposite directions with same ease, so that the system and surroundings pass through exactly the same intermediate state as in the direct process. e.g., An ideal gas allowed to expand slowly and then compressed slowly so as to reach its initial state. Irreversible Process: It is a process which can't be made to proceed in the reverse direction with the same ease and the system does not pass through the same intermediate states as in direct processes. e.g., Decay of organic matter, rusting of iron.
View full question & answer→Question 143 Marks
A certain gas at atmospheric pressure is compressed adiabatically so that its volume becomes half of its original volume. Calculate the resulting pressure in $Nm^{-2}$. Take $\gamma=1.4$ for air.
AnswerLet the original volume, $V_1 = V $
$\therefore$ Final valume, $\text{V}_2=\frac{\text{V}}{2}$ Initial pressure, $P_1 = 0.76$ metre of Hg column. Let $P_2$ be the fanal pressure after compression. As the change is adiabatic, $\therefore\text{P}_1\text{V}_1^\gamma=\text{P}_2\text{V}_2^\gamma$
$\text{P}_2=\text{P}_1\Big(\frac{\text{V}_1}{\text{V}_2}\Big)^\gamma=\text{P}_1\bigg(\frac{\text{V}}{\frac{\text{V}}2{}}\bigg)^{1.4}$
$\text{P}_2=0.76\times(2)^{1.4}$
$\text{P}_2=2.00$ metre of Hg column, As $\text{P}=\text{h}\rho\text{g}$
$\therefore\text{P}_22.00\times(13.6\times10^3)\times9.8\text{ Nm}^{-2}$
$\text{P}_2=2.672\times10^5\text{Nm}^{-2}$
View full question & answer→Question 153 Marks
Obtain an expression for work done by a gas in an isothermal expansion.
AnswerFor a small change in volume, work done is given by, dW = P dV We know, PV = nRT
$\therefore\text{P}=\frac{\text{nRT}}{\text{V}}$ For T = constant, $\text{dW}=\text{nRT}\frac{\text{dV}}{\text{d}}$ Net work done under isothermal condition to change the volume from $V_i$ to $V_f$ is, $\text{W}=\int\limits^{\text{V}_\text{f}}_{\text{V}_\text{i}}\text{dW}=\text{nRT}\int\limits^{\text{V}_\text{f}}_{\text{V}_\text{i}}\text{}\frac{\text{dV}}{\text{V}}$
$=\text{nRT}\Big|\log_\text{e}\text{V}\Big|^{\text{V}_\text{f}}_{\text{V}_\text{i}}$
$=\text{nRT}\log_\text{e}\Big(\frac{\text{V}_\text{f}}{\text{ V}_\text{i}}\Big)$
$\therefore\text{W}=2.3026\text{ nRT }\log_{10}\Big(\frac{\text{V}_\text{f}}{\text{V}_\text{i}}\Big)$ Where n is the number of moles. If $P_f$ and $P_i$ are the pressures, we can also write, $\text{W}=2.3026\text{ nRT }\log_{10}\Big(\frac{\text{P}_\text{i}}{\text{P}_\text{f}}\Big)$
View full question & answer→Question 163 Marks
A refrigerator has to transfer an average of 263J of heat per second from temperature -10°C to 25°C. Calculate the average power consumed assuming ideal reversible cycle and no other losses.
AnswerHere, $\text{T}_1=25+273$ $=298\text{K}$ $\text{T}_2=-10+273$ $=263\text{K}$ $\text{Q}_2=263\ \text{J}\text{s}^{-1}$ Since, $\frac{\text{Q}_1}{\text{Q}_2}=\frac{\text{T}_1}{\text{T}_2}$ $\Rightarrow\text{Q}=\frac{\text{T}_1}{\text{T}_2}\times\text{Q}_2$ $=\frac{298}{263}\times263=29 8\ \text{J}\text{s}^{-1}$ $\therefore$ Averege power consumed $=\text{Q}_1-\text{Q}_2$ $=(298-263)\ \text{J}\text{s}^{-1}=35\ \text{W}.$
View full question & answer→Question 173 Marks
Find out whether these phenomena are reversible or not :
- Waterfall.
- Rusting of iron.
Answer
- Waterfall: The falling of water cannot be reversible process. During the water fall, its potential energy convert into kinetic energy of the water.
On striking the ground, some part of potential energy converts into heat and $($sound not possible that heat and the sound$)$. In nature, it automatically convert the kinetic energy and potential energy so that the water will rise back so waterfall is not a reversible process.
- Rusting of iron: In rusting of iron, the iron become oxidised with the oxygen of the air as it is a chemical reaction, it cannot be reversed.
View full question & answer→Question 183 Marks
An ideal gas is taken through a cyclic thermodynamic process through four steps. The amount of heat involved in these steps are $Q_1=5960 \mathrm{~J}, Q_2=-5585 \mathrm{~J} . Q_3=-2980 \mathrm{~J}$ and $Q_4=3645$ respectively. The corresponding quantities of work involved are $W_1=200 \mathrm{~J}, \mathrm{~W}_2=-825 \mathrm{~J}, \mathrm{~W}_3=-1100 \mathrm{~J}$ and $\mathrm{W}_4$ respectively. Find the value of $\mathrm{W}_4$. What is the efficiency of the cycle?
AnswerAs the process is cyclic, therefore, $\Delta\text{U}=0$ According to first law of thermodynamics $\Delta\text{Q}=\Delta\text{U}+\Delta\text{W}=\Delta\text{W}$
$\Delta\text{W}=\Delta\text{Q}$ or $\text{W}_1+\text{W}_2+\text{W}_3+\text{W}_4$
$=\text{Q}_1+\text{Q}_2+\text{Q}_3+\text{Q}_4$
$\text{W}_4(\text{Q}_1+\text{Q}_2+\text{Q}_3+\text{Q}_4)-(\text{W}_1+\text{W}_2+\text{W}_3)$
$=(5960-5585-2980+3645)$
$-(-200-825-1100)$
$=1040-275=765\text{J}$
$\text{Effeciency}=\frac{\text{Net work done}}{\text{total heat absorbed}}$
$=\frac{\text{W}_1+\text{W}_2+\text{W}_3+\text{W}_4}{\text{Q}_1+\text{Q}_4}$
$\eta=\frac{2200-825-1100+765}{5960+3645}=\frac{1040}{9605}=0.1083$
$\eta=0.1083\times100\%=10.83\%$
View full question & answer→Question 193 Marks
What is meant by the term 'Molar specific heat of a gas? The molar specific heat of hydrogen in the temperature range of about 250K to 750K is about $\Big(\frac52\Big)\text{R}.$ At lower temperatures, the value of molar specific heat of hydrogen decreases to the value typical of monoatomic gases $\Big(\frac32\Big)\text{R}$ while at higher temperatures, it tends to the value $\Big(\frac72\Big)\text{R.}$ Explain.
AnswerMolar specific heat capacity of a gas refers to the amount of energy required for 1 mole of a substance to raise its temperature by 1K. In the temperature beyond 70K, rotational motion of $H_2$_ gas starts. So at 250K < T < 750K, the number of degrees of freedom becomes five- 2 rotational and 3 translational. $\therefore\text{C}_\text{V}=\frac{\text{f}}{2}\text{R}$ becomes $\text{C}_\text{V}=\frac52\text{R}$ For lower temperatures only translational degrees of freedom will exist and no rotational freedom. $\therefore\text{C}_\text{V}=\frac{3}{2}\text{R}$ At higher temperature vibrational motion of $H_2$ also starts. So at T > 7.50K, number of degrees of freedom are $\text{C}_\text{V}=\frac72\text{R}$
View full question & answer→Question 203 Marks
One mole of an ideal gas is taken in a Carnot engine working between $27°C$ and $227°C$. The useful work done in one cycle is $600J$. Calculate the ratio of the volume of the gas at the end and beginning of the isothermal expansion. Given $R = 8.31J mole^{-1} K^{-1}$.
AnswerHere, $T_2 = 27^\circ C = (27 + 273)K = 300K T_1 = 227^\circ C = (227 + 273)K = 500K W = 600J, R = 8.31J mole^{-1}K^{-1}$ As $\text{W}=2.303\text{R}(\text{T}_1-\text{T}_2)\log_{10}\frac{\text{V}_2}{\text{V}_1}$
$\therefore\log_{10}\frac{\text{V}_2}{\text{V}_1}=\frac{\text{W}}{2.303\text{R(T}_1-\text{T}_2)}$
$=\frac{600}{2.303\times8.31(500-300)}=0.1568$
$=\frac{\text{V}_2}{\text{V}_1}=\text{antilog}(0.1568)=1.435$
View full question & answer→Question 213 Marks
Define molar specific heat. Write its units.
AnswerMolar specific heat is defined as the amount of heat required to raise the temperature of one mole of a gas through 1K at constant volume or at constant pressure. If the volume is constant, it is called the molar specific heat at constant volume. Similarly, if the pressure is constant, it is called the molar specific heat at constant pressure. It is expressed in $\mathrm{J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ or $\mathrm{J} \mathrm{~mol}^{-1\circ} \mathrm{C}^{-1}$.
View full question & answer→Question 223 Marks
The heat of combustion of ethane gas is $373$ kcal. Per mole. Assuming that $50\%$ of heat is lost, how many litres of ethane measured at STP must be burnt to convert $50g$ of water at $10°C$ to steam at $100°C$? One mole of gas occupies $22.4$ litres at STP. Take latent heat of steam = $2.25 \times 10^6J/g^{-1}$.
AnswerTotal heat energy required to convert 50g of water at $10^\circ C$ to steam at $100^\circ C$, $=\text{cm}\Delta\text{T}+\text{mL}$
$=1000\times50\times(100-10)+\frac{50\times2.25\times10^6}{4.2}$
$=4.5\times10^6+26.79\times10^6$
$=31.29\times10^6\text{ cal}$ As 50% of heat is lost, $\therefore$ Total heat product = $2 \times 31.29 \times 10^6cal$ Heat of combustion = $373 \times 10^3cal/mole$
$\therefore$ No. of mole of ethane to be burnt, $=\frac{2\times31.29\times10^6}{373\times10^3}\text{mole}$ Volume of ethane, $=\frac{2\times31.29\times10^6}{373\times10^3}$
$=3758.2\text{ liters}$
View full question & answer→Question 233 Marks
Consider that an ideal gas ( $n$ moles) is expanding in a process given by $p=f(V)$, which passes through a point $\left(V_0\right.$, $\left.p_0\right)$. Show that the gas is absorbing heat at $\left(P_0, V_0\right)$, if the slope of the curve $p=f(V)$ is larger than the slope of the adiabat passing through $\left(\mathrm{P}_0, \mathrm{~V}_0\right)$.
AnswerSlope of $p=f(V)$, curve at $\left(V_0, p_0\right)=f\left(V_0\right)$ Slope of adiabat at $\left(V_0, p_0\right)=K(-\gamma) V_0^{-1-\gamma}=-\gamma \frac{p_0}{V_0}$
Now, heat absorbed in the proccess $p=f(V) d Q=d U+d W=n C_V d t+p d V$
Since, $\text{T}=\Big(\frac{1}{\text{nR}}\Big)\text{pV}=\Big(\frac1{\text{nR}}\Big)\text{V f(V)}$
$\therefore\text{dT}=\Big(\frac{1}{\text{nR}}\Big)[\text{f(V)}+\text{V f}'\text{(V)}]\text{dV}$
Thus, $\frac{\text{dQ}}{\text{dV}}\bigg|_{\text{V}=\text{V}_0}=\frac{\text{C}_\text{V}}{\text{R}}[\text{f(V}_0)+\text{f}'\text{(V)}]+\text{f}(\text{V}_0)$ $=\Big[\frac{1}{\gamma+1}+1\Big]\text{f}(\text{V}_0)+\frac{\text{V}_0\text{f}'\text{(V}_0)}{\gamma-1}$ $=\frac{\gamma}{\gamma-1}\text{p}_0+\frac{\text{V}_0}{\gamma-1}\text{f}'(\text{V}_0)$ Heat is absorbed when $\frac{\text{dQ}}{\text{dV}}>0$ when gas expands, that is when, $\gamma\text{p}_0+\text{V}_0\text{f}'(\text{V}_0)>0$ $\text{f}'(\text{V}_0)>-\gamma\frac{\text{p}_0}{\text{V}_0}$
View full question & answer→Question 243 Marks
How many grams of ice at -14°C are needed to cool 200 grams of water from 25°C to 10°C? Take specific heat of ice = 0.5 cal/ g/ °C and latent heat of ice = 80 cal/ g.
AnswerHeat extracted from water $\text{Q}_1=\text{cm}\Delta\text{T}$ $=1\times200(25-10)=3000\text{ cal}$ Heat absorbed by m gram of ice (at -14°C) to convert it to water at 10°C, $\text{Q}_2=(\text{cm}\Delta\text{T})_{\text{ice}}+\text{mL}+(\text{cm}\Delta\text{T})_{\text{water}}$ $=0.5\text{ m}\times14+\text{m}\times80+1\times\text{m}\times10$ $\text{Q}_2=97\text{m cal}$ As $\text{Q}_2=\text{Q}_1,$ $\therefore\text{97m}=3000$ $\Rightarrow\text{m}=31\text{g}$
View full question & answer→Question 253 Marks
The temperature of $3\ kg$ krypton gas is raised from $-29^\circ C$ to $89^\circ C$.
- If this is done at constant volume, compute the heat added, the work done, and the change in internal energy.
- Repeat if the heating process is at constant pressure.
For $K_r, \text{C}_\upsilon=0.0357\ \text{cal/gm}^\circ\text{C}$ and $\text{C}_\text{P}=0.0595\ \text{cal/gm}^\circ\text{C.}$ Answer
- According to first law of thermodynamics,
$\Delta\text{Q}=\Delta\text{U}+\Delta\text{W}$
At constant volume, $\Delta\text{W}=0$ so $\Delta\text{Q}=\Delta\text{U}$
- So heat added $\Delta\text{Q}=\Delta\text{U}=\text{mc}_\upsilon\Delta\text{T}$
$=(3\times10^3)0.0357\times100$
$=10710\text{cal}=1071\text{ kcal}$
- Work done $\Delta\text{W}=0$
- Change in internal energy $=10.71\times4.184=44.8\ \text{KJ}$
- At constant pressure, $\Delta\text{Q}=\text{mC}_\text{P}\Delta\text{T}$
$=(3\times10^3)0.0595\times10$
- Change in the internal energy will be the same i.e.,
$\Delta\text{U}=10.71\text{ kcal}$
Work done $\Delta\text{W}=\Delta\text{Q}-\Delta\text{U}$
$=17.85-10.71=7.14\ \text{ kcal}$
$=7.14\times4.184\text{kJ}=29.9\ \text{kJ}$
- The change in internal energy is the same as in isochoric process $= 44.8\ kJ.$
View full question & answer→Question 263 Marks
Prove that for an adiabatic process $\text{PV}^\gamma=\text{constant},$ where the symbols have their usual meanings.
AnswerFor an adiabatic process, $dQ = 0 dU = nC_vdT$ for a process, where there is a temperature change by dT. From gas equation, PV = nRT Differentiating both sides, we have, PdV + VdP = nRdT $\text{dT}=\frac{\text{PdV}+\text{VdP}}{\text{nR}}\dots(\text{ii})$ From first law of thermodynamics, $0 = nC_VdT + PdV$ ...(ii) Putting dT from (i) in (ii), we have, $\text{nC}_\text{V}\Big(\frac{\text{PdV}+\text{VdP}}{\text{nR}}\Big)+\text{PdV}=0$
$\text{C}_\text{V}(\text{PdV}+\text{VdP})+\text{RP dV}=0$
$\text{C}_\text{V}(\text{PdV}+\text{VdP})+(\text{C}_\text{P}-\text{C}_\text{V})\text{PdV}=0$
$[\because\text{R}=\text{C}_\text{P}-\text{C}_\text{V}]$
$\text{C}_\text{V}\text{VdP}+\text{C}_\text{P}\text{PdV}=0$ or $\frac{\text{dP}}{\text{P}}+\frac{\text{dV}}{\text{P}}\gamma=0$
$\Big[\because\frac{\text{C}_\text{P}}{\text{C}_\text{V}}=\gamma\Big]$ Integrating, we get, $\int\frac{\text{dP}}{\text{P}}+\gamma\int\frac{\text{dV}}{\text{V}}=\text{constant}$
$\log\text{P}+\gamma\log\text{V}=\text{constant}$
$\log\text{PV}^\gamma=\text{constant}$
$\text{PV}^\gamma=\text{constant}$ This is the equation for an adiabatic change in an ideal gas.
View full question & answer→Question 273 Marks
Absolute zero temperature is not the temperature of zero energy. Explain.
AnswerAt absolute zero, the energy of translatory motion of molecules ceases but the other forms of energy such as inter-molecular, potential energy of molecular motion, etc. do not become zero. Therefore, absolute temperature is not the temperature of zero energy.
View full question & answer→Question 283 Marks
$0.75g$ of petroleum was burnt in a bomb calorimeter which contains $2kg$ of water and has a water equivalent $500$ grams. The rise in temp. was $3°C$. Determine the calorific value of petroleum.
AnswerHeat absorbed by water, $\text{Q}_1=\text{cm}\Delta\text{T}=1\times2000\times3$
$=6\times10^3\text{cal}$ Heat absorbed by calorimeter, $\text{Q}_2=\text{W}\times\Delta\text{T}=500\times3$
$=1.5\times10^3\text{cal}$ Total heat produced = $Q_1 + Q_2 = 7.5 \times 10^3$ cal Calorific value of fuel $=\frac{7.5\times10^3}{0.75}=10^4\text{cal/g}$
View full question & answer→Question 293 Marks
An ideal engine works between temperatures $T_1$ and $T_2$. It derives an ideal refrigerator that works between temperatures $T_3$ and $T_4$. Find the ratio $\frac{\text{Q}_3}{\text{Q}_1}$ in terms of $T_1, T_2, T_3$ and $T_4$.
AnswerW = work done by engine = $Q_1 - Q_2$ and W = work done supplied to refrigerator = $Q_3 - Q_4 \Rightarrow\text{Q}_1-\text{Q}_2=\text{Q}_3-\text{Q}_4$ Dividing by $Q_1$, $1-\frac{\text{Q}_2}{\text{Q}_1}=\frac{\text{Q}_3}{\text{Q}_1}-\frac{\text{Q}_4}{\text{Q}_1}$
$\Rightarrow\frac{\text{Q}_3}{\text{Q}_1}=1-\frac{\text{Q}_2}{\text{Q}_1}+\frac{\text{Q}_4}{\text{Q}_1}$
$=1-\frac{\text{T}_2}{\text{T}_1}+\frac{\text{T}_4}{\text{T}_1}$
$=\frac{\text{T}_1-\text{T}_2+\text{T}_4}{\text{T}_1}$
View full question & answer→Question 303 Marks
Distinguish between a cyclic process and a non-cyclic process.
Answer A cyclic process is that in which the system returns to its initial state after under going a series of changes. A non-cyclic process is that in which the system does not return to its initial state.
View full question & answer→Question 313 Marks
One mole of an ideal gas undergoes a cyclic change ABCD. From the given diagram, calculate the net work done in the process. $1$ atmosphere = $10^\circ dyne\ cm^{-2}$.
AnswerIn a cyclic change, work done is equal to area of the loop ABCD.
As the loop is traced in clockwise direction, work done is positive. $W=$ area $A B C D=D C \times D A$
Now, $D C=4-1=3$ litre $=3 \times 10^3 \mathrm{~cm}^3 \mathrm{DA}=5-2=3 \mathrm{~atm}=3 \times 10^6$ dyne $\mathrm{cm}^{-2}$
$\therefore \mathrm{~W}=\mathrm{DC} \times \mathrm{DA}=3 \times 10^3 \times 3 \times 10^6=9 \times 10^9 \mathrm{erg}$. View full question & answer→Question 323 Marks
Derive an expression for the work done in an isothermal process.
AnswerFor a small change in volume, work done is given by, DW =P dV We, know, PV = nRT $\Rightarrow\text{P}=\frac{\text{nRT}}{\text{V}}$ For T = costant, $\text{dW}=\text{nRT}=\frac{\text{dV}}{\text{V}}$ Net work done under isothermal condition to change the valume from $V_i$ to $V_f$ is, $\text{W}=\int\limits^{\text{V}_\text{f}}_{\text{V}_\text{i}}\text{dW}=\text{nRT}\int\limits^{\text{V}_\text{f}}_{\text{V}_\text{i}}\frac{\text{dV}}{\text{V}}$
$=\text{nRT}\Big|\log_\text{e}\text{V}\Big|^{\text{V}_\text{f}}_{\text{V}_\text{i}}$
$\text{W}=\text{nRT}\log_\text{e}\Big(\frac{\text{V}_\text{f}}{\text{V}_\text{i}}\Big)$
$\therefore\text{W}=2.3026\text{ nRT }\log_{10}\Big(\frac{\text{V}_\text{f}}{\text{v}_i}\Big)$ Where n is the number of moles. If $P_f$ and $P_i$ are the pressures, we can also write, $\text{W}=2.3026\text{ nRT }\log_{10}\Big(\frac{\text{P}_\text{i}}{\text{P}_\text{i}}\Big)$
View full question & answer→Question 333 Marks
Two Carnot engines $A$ and $B$ are operated in series. The first one $A$ receives heat at $800K$ and rejects to a reservoir at temperature $T\ K$ The second engine $B$ receives the heat rejected by the first engine and in turn rejects to a heat reservoir at $300K$. Calculate the temperature $T\ K$ for the following cases.
- When the outputs of the two engines are equal.
- When the efficiencies of the two engines are equal.
AnswerFor engine A, $T_1 = 800K, T_2 = T K$ Effieciency, $\eta_\text{A}=1-\frac{\text{T}_2}{\text{T}_1}=1-\frac{\text{T}}{800}$
Also, $\frac{\text{Q}_2}{\text{Q}_1}=\frac{\text{T}_2}{\text{T}_1}=\frac{\text{T}}{800}$
Work output, $\text{W}_\text{A}=\text{Q}_1-\text{Q}_2=\eta_\text{A}\times\text{Q}_1$
$\Big[\because\eta_\text{A}=1-\frac{\text{Q}_2}{\text{Q}_1}\Big]$ Or $\text{W}_\text{a}=\Big(1-\frac{\text{T}}{800}\Big)\text{Q}_1$ For engine B, $\text{T}'_1=\text{T K},\text{ T}_2'=300\text{K}$
Efficiency, $\eta_\text{B}=1-\frac{\text{T}_2'}{\text{T}_1'}=1-\frac{300}{\text{T}}$
Work output, $\text{W}_\text{B}=\text{Q}'_1-\text{Q}_2'=\eta_\text{B}\times\text{Q}'_1=\Big(1-\frac{300}{\text{T}}\Big)\text{Q}'_1$
Since, the engine $B$ absorbs the heat rejected by the engine $A$,
so $\text{Q}_1'=\text{Q}_2\therefore\text{W}_\text{B}=\Big(1-\frac{300}{\text{T}}\Big)\text{Q}_2$
- When output of the two engins are equal,
$\text{W}_\text{A}=\text{W}_\text{B}$
$\Big(1-\frac{\text{T}}{800}\Big)\text{Q}_1=\Big(1-\frac{300}{\text{T}}\Big)\text{Q}_2$
$\Big(1-\frac{\text{T}}{800}\Big)=\Big(1-\frac{300}{\text{T}}\Big)\frac{\text{Q}_2}{\text{Q}_1}=\Big(1-\frac{300}{\text{T}}\Big)\frac{\text{T}}{800}$
On solving, we get $T = 550K$
- When the efficiencies are equal, $\eta_\text{A}=\eta_\text{B}$
$1-\frac{\text{T}}{800}=1-\frac{300}{\text{T}}$
$\text{T}^2=24\times10^4$
$\therefore\text{T}=489.9\text{K}$ View full question & answer→Question 343 Marks
An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
AnswerHeat is supplied to the system at a rate of 100W. $\therefore$ Heat supplied, Q = 100J/s The system performs at a rate of 75J/s. $\therefore$ Work done, W = 75J/s From the first law of thermodynamics, we have: Q = U + W Where, U = Internal energy $\therefore$ U = Q - W = 100 - 75 = 25J/s = 25W Therefore, the internal energy of the given electric heater increases at a rate of 25W.
View full question & answer→Question 353 Marks
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig.
Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.
AnswerTotal work done by the gas from D to E to F = Area of $\triangle \mathrm{DEF}$ Area of $\triangle \mathrm{DEF}=(1 / 2) \mathrm{DE} \times \mathrm{EF}$ Where, $\mathrm{DF}=$ Change in pressure $=600 \mathrm{~N} / \mathrm{m}^2-300 \mathrm{~N} / \mathrm{m}^2=300 \mathrm{~N} / \mathrm{m}^2 \mathrm{FE}=$ Change in volume $=5.0 \mathrm{~m}^3-2.0 \mathrm{~m}^3=3.0 \mathrm{~m}^3$ Area of $\Delta \mathrm{DEF}=(1 / 2) \times 300 \times 3=450$ J Therefore, the total work done by the gas from $D$ to $E$ to $F$ is 450 J.
View full question & answer→Question 363 Marks
Prove that the slope of P-V graph for an adiabatic process is $\gamma$ times that of the isothermal process.
AnswerFor isothermal process, PV = constantDifferentiating, VdP + PDV = 0
$\therefore\frac{\text{dP}}{\text{dV}}=-\frac{\text{P}}{\text{V}}$
For adiabatic process, $\text{PV}^{\gamma}=$ constant.
Differentiating,
$\text{V}^{\gamma}\text{dP}+\gamma\text{PV}^{\gamma-1}\text{dV}=0$
$\therefore\frac{\text{dP}}{\text{dV}}=-\frac{\gamma\text{P}}{\text{V}}$
Comparing the two ratios, we can say, slope of adiabatic process is $\gamma$ times the slope of isothermal process.
View full question & answer→Question 373 Marks
Calculate the specific heat capacity at constant volume for a gas. Given specific heat capacity at constant pressure is $6.85 \mathrm{cal} \mathrm{~mole}^{-1} \mathrm{~K}^{-2}, \mathrm{R}=8.31 \mathrm{~J} \mathrm{~mole}^{-1} \mathrm{~K}^{-1}$ and $\mathrm{J}=4.18 \mathrm{~J} \mathrm{cal}^{-1}$.
AnswerWe know that, $\text{C}_\text{p}-\text{C}_\text{V}=\frac{\text{R}}{\text{J}}$ $6.85-\text{C}_\text{V}=\frac{8.31}{4.18}$ $\Rightarrow6.85-\text{C}_\text{V}=1.988$ $\Rightarrow \text{C}_\text{V}=6.85-1.988$ $\Rightarrow\text{C}_\text{V}=4.862\text{ cal mole}^{-1}\text{K}^{-1}$
View full question & answer→Question 383 Marks
Explain why: Two bodies at different temperatures $T_1$ and $T_2$ if brought in thermal contact do not necessarily settle to the mean temperature $(T_1 + T_2 )/2$.
AnswerWhen two bodies at different temperatures $T_1$ and $T_2$ are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature $(T_1 + T_2)/2$ only when the thermal capacities of both the bodies are equal.
View full question & answer→Question 393 Marks
What is the coefficient of performance $(\beta)$ of a Carnot refrigerator working between 30°C and 0°C?
AnswerHere $\text{T}_2=0^\circ\text{C}=273\text{K}$ $\text{T}_1=30^\circ\text{C}$ $=273+30=303\text{K}$ $\beta=?$ Using the relation, $\beta=\frac{\text{T}_2}{\text{T}_1\text{T}_2},$ we get $\beta=\frac{273}{303-273}$ $=\frac{273}{30}=9.1.$
View full question & answer→Question 403 Marks
A refrigerator is to maintain eatables kept inside at $9°C$. If room temperature is $36°C$, calculate the coefficient of performance.
AnswerTemperature inside the refrigerator, $\mathrm{T}_1=9^{\circ} \mathrm{C}=282 \mathrm{~K}$ Room temperature, $\mathrm{T}_2=36^{\circ} \mathrm{C}=309 \mathrm{~K}$ Coefficient of performance $=T_1 / T_2-T_1=282 / 309-282=282 / 27=10.44$ Therefore, the coefficient of performance of the given refrigerator is 10.44 .
View full question & answer→Question 413 Marks
200J of work is done on a gas to reduce its volume by compressing it. If this change is done under adiabatic conditions, find out the change in internal energy of the gas and also the amount of heat absorbed by the gas.
AnswerIn adiabatic changes, dQ = 0 $\therefore\text{dQ}=\text{dU}+\text{dW}=0$ $\text{dU}=-\text{dW}=-(-200\text{J})=200\text{J}$ Internal energy increases by 200J. Heat absorbed is zero.
View full question & answer→Question 423 Marks
An ideal refrigerator is working between the temperature of ice and temperature of atmosphere at $300K$. Find the energy which has been supplied to it to freeze $2kg$ of water at $0°C$. Given that latent heat of ice $3.33 × 105J/ kg$.
AnswerHere, $T_1=300 \mathrm{~K}, \mathrm{~T}_2=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ Heat etracted, $\mathrm{Q}_2=\mathrm{mL} 1=2 \mathrm{~kg} \times 3.33 \times 10^5 \mathrm{~J} / \mathrm{kg}=6.66 \times 10^5 \mathrm{~J}$ As, $\beta=\frac{\text{Q}_2}{\text{W}}=\frac{\text{T}_2}{\text{T}_1-\text{T}_2}$ $\therefore\text{W}=\frac{\text{Q}_2(\text{T}_1-\text{T}_2)}{\text{T}_2}$ $=\frac{6.66\times10^5\times(300-273}{273}$ $=65868\text{J}\simeq6.5\times10^4\text{J}$
View full question & answer→Question 433 Marks
A steam engine delivers $5.4 \times 10^8 \mathrm{~J}$ of work per minute and services $3.6 \times 10^9 \mathrm{~J}$ of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
AnswerWork done by the steam engine per minute, $\mathrm{W}=5.4 \times 10^8 \mathrm{~J}$ Heat supplied from the boiler, $\mathrm{H}=3.6 \times 10^9 \mathrm{~J}$ Efficiency of the engine $=$ output energy/Input energy
$\therefore \mathrm{n}=\mathrm{W} / \mathrm{H}=5.4 \times 10^8 / 3.6 \times 10^9$
Hence, the percentage efficiency of the engine is $15 \%$. Amount of heat wasted $=3.6 \times 10^9-5.4 \times 10^8=30.6 \times 10^8=3.06 \times 10^9 \mathrm{~J}$
Therefore, the amount of heat wasted per minute is $3.06 \times 10^9 \mathrm{~J}$.
View full question & answer→Question 443 Marks
Is it possible to increase the temperature of a gas without adding heat to it? Explain.
AnswerYes, it is possible to increase the temperature of a gas without adding heat to it, during adiabatic compression the temperature of a gas increases while no heat is given to it. For an adiabatic compression, no heat is given or taken out in adiabatic process. Therefore,$\Delta\text{Q}=0$ According to the first law of thermodynamics, $\Delta\text{Q}=\Delta\text{U}+\Delta\text{W}$ $\Delta\text{U}=-\Delta\text{W}(\Delta\text{Q}=0)$ In compression work is done on the gas, i.e. work done is negativ. Therefore, $\Delta\text{U}=$ positive. Hence, internal energy of the gas increases due to which its temperature increases.
View full question & answer→Question 453 Marks
A car tyre contains air at a pressure of $4$ atm and its temperature is $27°C$. The tyre suddenly bursts. Calculate the resulting temperature. $(\gamma=1.4)$
AnswerP_1 = 4atm, P_2 = 1atm, $T_1 -27^\circ C = 300K$ and $\gamma=1.4$ The sudden burst of tyre is an adiabatic process, in which , $\text{P}^{1-\gamma}_1\text{T}^\gamma_2=\text{P}^{1-\gamma}_2\text{T}^{\gamma}_2$
$\therefore\text{T}_2=\text{T}_1\Big(\frac{\text{P}_1}{\text{P}_2}\Big)^{\frac{1-\gamma}{\gamma}}=\text{T}_1\Big(\frac{\text{P}_2}{\text{P}_1}\Big)^{\frac{\gamma-1}{\gamma}}$
$=300\Big(\frac{4}{1}\Big)^{\frac{1.4-1}{1.4}}=201.9$
$=202\text{K}$ or $-71^\circ\text{C}.$
View full question & answer→Question 463 Marks
Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in Fig. 
Given the internal energy for one mole of gas at temperature T is $(3/2)$ RT, find the heat supplied to the gas when it is taken from state (1) to (2) with $V_2 = 2V_1$. Answer$\therefore\ \text{PV}=$ constant = K (given) or $\text{P}_1\text{V}_1^\frac{1}{2}=\text{P}_2\text{V}_2^\frac{1}{2}= \text{K}\ \text{and}\ \text{P}=\frac{\text{K}}{\text{V}^\frac{1}{2}}$ Given that internal energy U of gas is $\text{U}=\Big(\frac{3}{2}\Big)\text{RT}$
$\Delta\text{U}=\frac{3}{2}\text{RdT}=\frac{3}{2}\text{R}(\text{T}_2-\text{T}_1)$
$\therefore\ \text{T}_2=\sqrt2\text{T}_1,$ from part (b) $\Delta\text{U}=\frac{3}{2}\text{R}\big[\sqrt2\text{T}_1-\text{T}_1\big]=\frac{3}{2}\text{R}\text{T}_1\big(\sqrt2-1\big)$ Form part (a) $\text{dW}=2\text{P}_1\text{V}^\frac{1}{2}\big(\sqrt{\text{V}_2}-\sqrt{\text{V}_1}\big)$
$\because\ \text{V}_2=2\text{V}_1$ (given) so, $\sqrt{\text{V}_2}=\sqrt2\sqrt{\text{V}_1}$$$ then $\text{dW}=2\text{P}_1\text{V}_1^\frac{1}{2}\big(\sqrt2\sqrt{\text{V}_1}-\sqrt{\text{V}_1}\big)$
$=2\text{P}_1\text{V}_1\sqrt{\text{V}_1}\big[\sqrt2-1\big]$
$\text{dW}=2\text{P}_1\text{V}_1\big(\sqrt2-1\big)$
$\text{dW}=2\text{n}\text{R}\text{T}_1\big(\sqrt2-1\big)$
$\big(\therefore\ \text{P}_1\text{V}_1=\text{n}\text{R}\text{T}_1\big)$
$\therefore\ \text{n}=1\therefore\ \text{dW}=2\text{R}\text{T}_1\big(\sqrt2-1\big)$
$\therefore\ \text{dQ}=\text{dW}+\text{dU}=2\text{R}\text{T}_1\big(\sqrt2-1\big)+\frac{3}{2}\text{R}\text{T}_1\big(\sqrt2-1\big)$
$=\big(\sqrt2-1\big)\text{R}\text{T}_1\Big[2+\frac{3}{2}\Big]$
$\text{dQ}=-\big(\sqrt2-1\big)\text{RT} .$
View full question & answer→Question 473 Marks
Find the value of $C_v$ and $C_p$ for nitrogen (given $R = 8.3J mole^{-1} K^{-1}$, also for a diatomic gas, $\text{C}_\text{V}=\frac52\text{R}.$
AnswerAs nirogen is a diatomic molecule, $\therefore\ \text{C}_\text{V}=\frac{5}{2}\text{R}$
$=\frac{5}{2}\times8.3\ \text{J}\ \text{mol}\ ^{-1}\text{K}^{-1}$
$=20.75\ \text{J}\ \text{mol}^{-1}\text{K}^{-1}$ But $\text{C}_\text{P}-\text{C}_\text{V}=\text{R}$
$\therefore\text{C}_\text{P}=\text{C}_\text{V}+\text{R}$
$=(20.75+8.3)\ \text{J}\ \text{mol}^{-1}\text{K}^{-1}$
$=29.05\ \text{J}\ \text{mol}^{-1}\text{K}^{-1}$
View full question & answer→Question 483 Marks
The efficiency of a Carnot engine is $\frac12.$ If the sink temperature is reduced by $100°C$, then engine efficiency becomes $\frac23.$ Find
- Sink temperature.
- Source temperature.
- Explain, why a Carnot engine cannot have $100\%$ efficiency?
Answer
- Efficiency, $\eta=1-\frac{\text{T}_2}{\text{T}_1}$
Where, $T_2 =$ sink temperature
$T_1 =$ source temperature
$1-\frac{\text{T}_2-100}{\text{T}_1}\dots\text{(i)}$
$1-\Big(\frac{\text{T}_2-100}{\text{T}_1}\Big)=\frac23\dots\text{(ii)}$
From equation $(i), \frac{\text{T}_2}{\text{T}_1}=\frac{1}{2}$ and equation $(ii),$
${\text{T}_2-100\over\text{T}_1}=\frac32$
$\Rightarrow\text{T}_2=300\text{K}$
- Substituting in equation $(i), T_1 = 600K$
- As efficiency, $\eta_2$
- $\Rightarrow1-\frac{\text{T}_2}{\text{T}_1}$
$\therefore$ It equals to $1$ only when $\frac{\text{T}_2}{\text{T}_1}=0$ or $\text{T}_2=0\text{K}$
But absolute zero is not possible. View full question & answer→Question 493 Marks
An ideal gas changes its state from $L$ to $M$ by two path $\text{LNM}$ and $LM$.
- Is the work done same for two paths?
- The internal energy of gas at $L$ is $20J$ and the amount of heat needed to change its state through $LM$ is $400J$. What is the internal energy of gas at $M$?
Answer
- $\text{W}_{\text{LN}}=\text{PdV}=0$
$\text{W}_{\text{NM}}=\text{P}[\text{V}_\text{M}-\text{V}_\text{M}]=10[6-2]=40\text{J}$
$\text{W}_\text{LMN}=\text{W}_{\text{NM}}=0+40=40\text{J}$
Along $\ce{LM \ W_{LM}} =$ Area under the curve $LM$
= Area of $\Delta\text{LMQ}+$ Area of rectangle $\text{LQZP.}$
$=\frac12\times\text{LQ}\times\text{MQ}+\text{LP}\times\text{PZ}$
$=\frac12\times4\times5+5\times4$
$=10+20=30\text{J}$
So work done is less along $LM.$
- $\text{U}_\text{L}=20\text{J}$
$\Delta\text{Q}=400\text{J}$
$\text{dQ}=\text{dU}+\text{dW}$
$=(\text{U}_\text{M}-\text{U}_\text{L})+\Delta\text{W}_{\text{LM}}$
$\text{U}_\text{M}=\text{dQ}+\text{U}_\text{L}-\Delta\text{W}_\text{LM}$
$=400+20-30$
$=390\text{J}$ View full question & answer→Question 503 Marks
State the first law of thermodynamics. List the sign conventions used in the energy dealt by the law.
AnswerAccording to the first law of thermodynamics, the total heat energy change dQ is the sum of the internal energy change dU and work done dW, i.e., dQ = dU + dW. Heat energy given to the system is +ve, taken out is -ve. Internal energy change is +ve with increase in temperature. Work done is +ve if volume increases and -ve if volume decreases.
View full question & answer→Question 513 Marks
State first law of thermodynamics. Why $C_p > C_v$? Prove $C_p - C_v = R$.
AnswerFirst Law of Thermodynamics: It is based on the law of conservation of energy. The total heat energy change in any system is the sum of the internal energy change and the work done, i.e. dQ = dU + dW. where dU is the internal energy change and dW = PdV is the work done by/ on the system. $C_p$ is greater because under constant pressure process, the energy also does work. Poof of $C_p - C_v = R$ Suppose one mole of a gas is heated at constant volume so that its temperature rises by dT. Heat supplied = $1 × C_V \times dT = C_VsdT$ ...(i) Since the volume is constant, the gas will not perform external work in accordance with the first law of thermodynamics and the heat supplied will be just equal to the increase in the internal energy of the gas.
$\therefore dU = C_VdT$ ...(ii) Let the gas be heated at constant pressure to increase its temperature by dT, and dQ be the amount of heat supplied, therefore, $dQ = 1 \times C_P \times dT = C_PdT$ ...(iii) The heat supplied at a constant pressure increases the temperature by dT, hence increases its internal energy by $dU = C_vdT$ as well as enables the gas to perform work dW. dW = PDV ...(iv) From the first law of thermodynamics, we have, dQ = dU + dW Substituting the values, we get, $C_PdT = C_VdT + PdV$ But PV = RT (For one mole of an ideal gas) or PdV = RdT,
$\therefore\text{C}_\text{P}\text{dt}=\text{C}_\text{P}\text{dT}+\text{RdT} C_P - C_V = R$ ...(v) This is the relationship between two principal specific heats of the gas when $C_p, C_v$ and R are measured in the units of either heat or of work.
View full question & answer→Question 523 Marks
Temperatures of the hot and cold reservoirs of a Carnot engine is raised by equal amounts. How the efficiency of the Carnot engine affected?
AnswerLet the initial temperatures of hot and cold reservoirs were $T_1$ and $T_2$. The efficiency of the Carnot engine is given by, So, intially $\eta=\frac{\text{T}_1-\text{T}_2}{\text{T}_1}\dots\text{(i)}$ As given the temperature of both the reservoirs is raised by equal amout t so $\text{T}'_1=\text{T}+\text{t}$ and $\text{T}'_2=\text{T}_2+\text{t}.$ The final efficiency of the Carnot engine will be $\eta'=\frac{\text{T}'_1-\text{T}_2'}{\text{T}'_1}$
$=\frac{(\text{T}_1+\text{t})-(\text{T}_2'+\text{t})}{\text{(T}_1+\text{t)}}$
$=\frac{\text{T}_1-\text{T}_2}{(\text{T}_1+\text{t})}\dots\text{(ii)}$ Dividing equation (ii) by equation (i), we have $\frac{\eta'}{\eta}=\frac{\Big(\frac{\text{T}_1-\text{T}_2}{\text{T}_1+\text{t}}\Big)}{\Big(\frac{\text{T}_1-\text{T}_2}{\text{T}_1}\Big){}}=\frac{\text{T}_1}{\text{T}_1+\text{t}}\dots\text{(iii)}$ As $\eta'<\eta,$ i.e. the effieciency of Cornot engine decreases.
View full question & answer→Question 533 Marks
A perfect Carnot engine utilizes an ideal gas. The source temperature is $500K$ and sink temperature is $375K$. If the engine takes $600K$ cal per cycle from the source, compute:
- The efficiency of the engine.
- Work done per cycle.
- Heat rejected to the sink per cycle.
AnswerHere $T_1 = 500K$ $T_2 = 375K Q_1 =$ Heat absorbed per cycle $= 600\ kcal$
-
$\therefore$ Using the relation,
$\eta=1-\frac{\text{T}_2}{\text{T}_1},$ we get
$\eta=\frac{\text{T}_1-\text{T}_2}{\text{T}_1}$
$=\frac{500-375}{500}$
$=\frac{125}{500}=0.25$
$\eta\%=0.25\times100$
$=25\%$
- Let $W =$ work done per cycle
$\therefore$ Using the relation
$\eta=\frac{\text{W}}{\text{Q}_1},$ we get
$\text{W}=\eta\text{Q}_1=0.25\times600\text{ kcal}=150\text{ kcal}$
$=150\times10^3\times4.2\text{J}$
$=6.3\times10^5\text{J}$
- Let $Q_2 =$ Heat rejected to the sink
$\therefore$ Using the relation
$\text{W}=\text{Q}_1-\text{Q}_2,$ we get
$\text{Q}_2=\text{Q}_1-\text{W}$
$=600-150=450\text{ kcal}$ View full question & answer→Question 543 Marks
A Carnot's engine whose sink is at a temperature of 300K has an efficiency of 40%. By how much should the temperature of the source be increased so as to increase the efficiency to 60%?
AnswerLet T be the temperature of the source. $\frac{40}{100}=\frac{\text{T}-300}{\text{T}}$ or $\frac25=\frac{\text{T}-300}{\text{T}}$ T = 500K Let the temperature be increased by $\theta,$ therefore $\frac{60}{100}=\frac{(\text{T}+\theta)-300}{(\text{T}+\theta)}$ $\frac35=\frac{500+\theta-300}{500+\theta}$ $\theta=\frac{500}2=250\text{K}$
View full question & answer→Question 553 Marks
Consider a Carnot's cycle operating between $T_1=500 \mathrm{~K}$ and $T_2=300 \mathrm{~K}$ producing 1 kJ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.
AnswerEfficiency of Carnot's engine $\mu=1-\frac{\text{T}_{2}}{\text{T}_{1}}$ Tempreature of source or reservior = $T_1 = 500K $
Tempreature of sink $T_2 = 300K$
$\therefore\mu=1-\frac{\text{T}_{2}}{\text{T}_{1}}$
$\frac{\text{Output work}}{\text{Input work}(E)}=1-\frac{300}{500}$
$\frac{1000\text{J}}{\text{x}}=1-0.6$
$\text{x}=\frac{1000}{\text{x}}=0.4$
$\text{x}=\frac{1000}{0.4}=2500 \text{J} .$
View full question & answer→Question 563 Marks
What are isotherms? Explain the process of liquiefication using it.
AnswerIsotherms are P-V graphs drawn at same temperature. When pressure is increased, volume is reduced at constant temperature. Beyond certain level, the pressure remains same for certain reduction in volume and then the pressure increases sharply. As temperature is increased, the constant portion reduces and at a temperature called critical temperature, the flat portion is absent. In each isotherm, before the flat portion is reached, the substance is in gaseous state and beyond the flat portion, it is in the liquid state. During the flat portion of P-V graph, the gas is under liquefication and so both gaseous and liquid states co-exist. Beyond T. it is not possible to liquefy the gas whatever large the pressure may be.
View full question & answer→Question 573 Marks
During India-Pakistan war, a soldier discovered that his lead bullet just melted when stopped by an obstacle. Calculate the velocity of the bullet if its temp. was $47.6^{\circ} \mathrm{C}$. Given: melting point of lead $=327^{\circ} \mathrm{C}$. Specific heat of lead $=0.03 \mathrm{cal} \mathrm{g}^{-10} \mathrm{C}^{-1}$, latent heat of fusion of lead $=6 \mathrm{cal} \mathrm{g}^{-1}$ and $\mathrm{J}=4.2 \times 10^7 \mathrm{erg} \mathrm{~car}^{-1}$. Assume that no heat is lost.
AnswerIncrease in temperature, $\theta=(327-47.6)^\circ\text{C}=279.4^\circ\text{C}$ Let m be the mass of the bullet. Heat required, $\text{Q}=\text{mS}\theta+\text{mL}$ or $\text{Q}=\text{m}(\text{S}\theta+\text{L})$ $=\text{m}(0.03\times279.4+6)$ $=14.38\text{m cal}$ Work done, $\text{W}=\frac12\text{mv}^2$ Where v is the velocity of bullet, Now, $\text{W}=\text{JQ}$ $\therefore\frac12\text{mv}^2=4.2\times10^7\times14.38\text{m}$ $\text{v}^2=2\times4.2\times10^7\times14.38$ $\text{v}=3.48\times10^4\text{cm s}^{-1}$
View full question & answer→Question 583 Marks
In a perfect Carnot's engine, the temperature of the source and the sink are $500K$ and $375K$ respectively. If the engine consumes $6 \times 10^5 cal$ per cycle, find $(i)$ the efficiency of the engine, $(ii)$ work done per cycle, and $(iii)$ the heat rejected to the sink per cycle.
Answer
- $\eta=\Big(1-\frac{\text{T}_2}{\text{T}_1}\Big)\times100$
$=\Big(1-\frac{375}{500}\Big)\times100=25\%$
- $\eta=\frac{\text{W}}{\text{Q}_1};$
$\text{W}=\eta\text{ Q}_1;\text{Q}_1=6000\times10^3\text{cal}$
$\text{W}=\frac{25}{100}\times600\times10^3\text{cal}$
$=150\times10^3\text{cal}=150\times10^3\times4.2\text{J}$
$=6.8\times10^5J($approx$)$
- $\text{W}=\text{Q}_1-\text{Q}_2;\text{Q}_2=\text{Q}_1-\text{W}$
$=600\times10^3-150\times10^3$
$=45\times10^4$ View full question & answer→Question 593 Marks
A Carnot engine absorbs 1000J of heat energy from a reservoir at 12.7°C and rejects 600J of heat energy during each cycle. Calculate. (i) efficiency of the engine, (ii) temperature of sink, (iii) amount of useful work done per cycle.
AnswerHere, $\text{Q}-1=1000\text{J},\text{Q}_2=600\text{J},$ $\text{T}_1=127^\circ\text{C}=127+273=400\text{K}$ $\eta=?,\ \text{T}_2=?,\ \text{W}=?$ From $\frac{\text{Q}_2}{\text{Q}_1}=\frac{\text{T}_2}{\text{T}_2},\text{T}_2=\frac{\text{Q}_2}{\text{Q}_1}\times\text{T}_1$ $\text{T}_2=\frac{600}{1000}\times400=240\text{K}$ $=240-273=-33^\circ\text{C}$ $\eta=1-\frac{\text{T}_2}{\text{T}_1}=1-\frac{240}{400}=0.4=40\%$ Also, $\text{W}=\text{Q}_1-\text{Q}_2=1000-600=400\text{J}$
View full question & answer→Question 603 Marks
What is the Internal energy in the process of vapourisation?
AnswerDuring vapourisation, volume increases. So work done $=P\left(V_f-V_i\right)$ and temperature does not change. So, $d Q=m L_v$
$\therefore$ from first law of thermodynamics, $d U=d Q-D W=m L_v-P\left(V_f-V_i\right)$.
View full question & answer→Question 613 Marks
Show that an adiabatic curve is always steaper than an isothermal curve.
AnswerFor Isothermal process, since PV = constant$\frac{\text{dP}}{\text{P}}=\frac{\text{dV}}{\text{V}}$
But for an adiabatic process, since
$\text{PV}^\gamma=\text{constant},\frac{\text{dP}}{\text{P}}=-\gamma\frac{\text{dV}}{\text{V}}$
So P-V graph is steaper for adiabatic process.
View full question & answer→Question 623 Marks
Two thermometers are constructed in the same way except that one has a spherical bulb and the other has an elongated cylindrical bulb. Which one will respond quickly to the temperature changes? Why?
AnswerThermometer with the cylindrical bulb. It is because the area of a cylinderical surface is greater than a spherical surface. Therefore, mercury in the cylindrical bulb reaches the temperature of the surrounding earlier.
View full question & answer→Question 633 Marks
State law of equi-partition of energy. Use this law to calculate specific heats of monoatomic, diatomic and triatomic gases.
AnswerAccording to the equi partition of energy, each degree of freedom will contribute an equal energy of $\frac12\text{R}$ per mole. In mono, di and triatomic gases hawing 3, 5 and 6 or 7 degrees of freedom, the internal energy will be $\frac32\text{R},\frac52\text{RT}$ and $\frac62\text{RT}$ or $\frac72\text{RT}.$ Using dU = $nC_VdT$ for mole, we get $\text{C}_\text{V}=\frac32\text{R},\frac{5}{2}\text{R},$ and 3R or $\frac72\text{R}$ for the three gases respectively.
View full question & answer→Question 643 Marks
What are 'Super heated water' and 'Super cooled vapour'?
AnswerWater in liquid phase at a temperature above 100°C and a pressure more than 1atm is called as super heated water. In a pressure cooker, water is heated at a pressure more than 1atm and temperature above 100°C. Steam below temperature 100°C is called super-cooled vapour.
View full question & answer→Question 653 Marks
Let the temperatures $T_1$ and $T_2$ of the two heat reservoirs in an ideal Carnot engine be $1500^{\circ} \mathrm{C}$ and $500^{\circ} \mathrm{C}$ respectively. Which of these, increasing $T_1$ by $100^{\circ} \mathrm{C}$ or decreasing $T_2$ by $100^{\circ} \mathrm{C}$, would result in a greater improvement in the efficiency of the engine?
AnswerThe efficiency of Carnot engine is given by, $\eta=1-\frac{\text{T}_2}{\text{T}_1}=\frac{\text{T}_1-\text{T}_2}{\text{T}_1}$
- When $T,$ is increased from $1500^\circ C$ to $1600^\circ C$ or $1600 + 273 = 1873K$ and $T,$ remains constant i.e., $500^\circ C$, or $500 + 273 = 773K$, then
$\eta_1=\frac{1873-773}{1873}=\frac{1100}{1873}$
$=\frac{1100\times100}{1873}=58.73\%$
- When $T_1$ remains constant i.e., $1500^\circ C$ or $1500 + 273 = 1773K$ and $T_2$ is decreased by $100 ^\circ C$ i.e., from $500^\circ C$ to $400^\circ C$ or $400 + 273 = 673K$ then
$\eta_2=\frac{1773-673}{1773}=\frac{1100}{1773}$
$=\frac{1100\times100}{1773}=62.04\%$
Thus, $\eta_2>\eta_1$ Hence efficiency will be increased if $T_2$ is decreased from $500^\circ C$ to $400^\circ C.$ View full question & answer→Question 663 Marks
No real engine can have efficiency greater than that of a Carnot engine working between the same two temperatures, why?
AnswerA Carnot engine is an ideal heat engine from the following points of view:
- There is absolutely no friction between the walls of cylinder and the piston.
- The working substance is an ideal gas. In a real engine, these conditions cannot be fulfilled and hence no heat engine working between the same two temperatures can have efficiency greater than that of Carnot engine.
View full question & answer→Question 673 Marks
Calculate the efficiency of a Carnot's engine working between steam point and ice point.
AnswerHere, steam point, $\text{T}_1=100\ ^\circ\text{C}$ $=100+273=373\text{K}$ Ice point , $\text{T}_2=0^\circ{\text{C}}$ $=0+273=273\text{K}$ As $\eta=1-\frac{\text{T}_2}{\text{T}_1}$ $\therefore\ \eta=1-\frac{273}{373}=\frac{100}{373}$ $=\frac{100}{373}\times100\%$ $=26.81\%$
View full question & answer→Question 683 Marks
The temperature of equal masses of three different liquids $\mathrm{A}, \mathrm{B}$ and C are $12^{\circ} \mathrm{C}, 19^{\circ} \mathrm{C}$ and $28^{\circ} \mathrm{C}$ respectively. The temperature when A and B are mixed is $16^{\circ} \mathrm{C}$. When B and C are mixed, the temperature is $23^{\circ} \mathrm{C}$. What would be the temperature when A and C are mixed?
AnswerLet $S_A, S_B, S_C$ be the specific heats of liquids A, B and C respectively. When A and B are mixed, $\text{mS}_{\text{A}}(16-12)=\text{mS}_\text{B}(19-16)$
$\text{S}_\text{A}=\frac34\text{S}_\text{B}\dots\text{(i)}$ When B and C are mixed, $\text{ms}_\text{B}(23-19)=\text{mS}_\text{C}(28-23)$
$\text{S}_\text{B}=\frac{5}{4}\text{S}_\text{C}\dots\text{(ii)}$ From (i) and (ii), $\text{S}_\text{A}=\frac{15}{16}\text{S}_\text{C}$ Substituting in (iii), we get $\frac{15}{16}(\theta-12)\text{S}_\text{C}=(28-\theta)\text{S}_\text{C}$ or $\theta=22.0^\circ\text{C}.$
View full question & answer→Question 693 Marks
What is a heat engine? What is the best way to increase efficiency of a heat engine? Is it possible to design a thermal engine that has $100\%$ efficiency?
AnswerA heat engine is a device (or a combination) which converts heat into work. Its efficiency, $\eta=\frac{\text{Work output}}{\text{Heat input}}$
$\eta=1-\frac{\text{T}_2}{\text{T}_1}$ Where, $T_2$ = temperature of sink $T_1$ = temperature of source. From above expression, we can see that for 100% efficiency, $T_2 = 0$ It is impossible to design a thermal engine that has 100% efficiency because it is not possible to have a sink with kelvin temperature.
View full question & answer→Question 703 Marks
- Describe a Carnot's cycle.
- A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in below figure. Its volume is then reduced to the original value from $E$ to $F$ by an isobaric process. Calculate the work done by the gas from $D$ to $E$ to $F$.
Answer
- Sadi Carnot devised an ideal cycle of operation of heat engine, which came to be known as Carnot cycle which is a set of four devices- a source at a high temperature (say $T_1$), a sink at a low temperature (say $T_2$), a non-conducting base and a cylinder with a working substance (a perfect gas) frictionless piston made of conducting base and non-conducting walls and piston.
- Change in pressure, $dP = 5.0 - 2.0\ atm = 3.0 \times 10^5 Nm^{-2}$
Change in volume $dV = 600 - 300 = 300\ cc = 300 \times 10^{-6} m^3$
Work done by the gas from D to E to F
= Area of $\Delta\text{DEF}$
$=\frac{1}{2}\times\text{dP}\times\text{dV}$
$=\frac12\times3.0\times10^5\times300\times10^{-6}\text{m}^3$
$=45\times10^6\times10^{-6}=45\text{J}$
View full question & answer→Question 713 Marks
A Carnot engine whose heat sink is at $27°C$ has an efficiency of $40\%$. By how many degrees should the temperature of source be changed to increase the efficiency by $10\%$ of the original efficiency?
AnswerHere, $T_2 = 27^\circ C = 27 + 273 = 300K \eta=40\%,\text{ T}_2=?$ From, $\eta=1-\frac{\text{T}_2}{\text{T}_1}$
$\frac{\text{T}_2}{\text{T}_1}=1-\eta$
$=1-\frac{40}{100}$
$=\frac{60}{100}$
$=\frac35$
$\text{T}_1=\frac53\text{t}_2$
$=\frac{5}{3}\times300=500\text{K}$ Increase in efficiency = 10% of 40 = 4% Let $\text{T}_1'$ be the new temprature of the source. As $\eta'=1-\frac{\text{T}_2}{\text{T}'_1},$
$\therefore\frac{\text{T}_2}{\text{T}'_1}=1-\eta'$
$=1-\frac{44}{100}$
$=\frac{56}{100}$
$\text{T}'_1=\frac{100}{56}$
$\text{T}_2=\frac{100}{56}\times300$
$=535.7\text{K}$
$\therefore$ Increase in temperature of source = 535.7 - 500 = 35.7K
View full question & answer→Question 723 Marks
A gram molecule of a gas at $127°C$ expands isothermally until its volume is doubled. Find the amount of work done and heat absorbed.
AnswerHere, temperature of the gas, T=273 + 127 = 400 K Let initial volume of the gas, $V_1 = V $
$\therefore$ Final volume of the gas, $V_2 = 2V$ In an isothermal expansion, Work done (W) $=2.3026\text{ RT }\log_{10}\frac{\text{V}_2}{\text{V}_1}$
$=2.3026\times8.3\times400\times\log_{10}\frac{2\text{V}}{\text{V}}$
$=2.3026\times8.3\times400\times0.3010$
$\text{W}=2.30\times10^{3}\text{joule}$ If H is the amount of heat absorbed, then $\text{H}=\frac{\text{W}}{\text{J}}=\frac{2.30\times10^3}{4.2}=548\text{b cal}$
View full question & answer→Question 733 Marks
The efficiency of heat engine cannot be $100\%$. Explain?
AnswerThe efficiency of heat engine is given by, $\eta=1-\frac{\text{Q}_2}{\text{Q}_1}$ $\text{Q}_2=$ Heat rejected to the sink, $\text{Q}_1=$ Heat absorbed from the source $\eta=1$ or 100% if and only if $Q_2 = 0$ This cannot happen because if $Q_2 = 0$, then the temperature of the working substance will go on increasing. A stage will come when the temperature of the working substance becomes equal to the temperature of the source. In this situation, there is no transfer of heat from source to the working substance. Hence, we will not get the output.
View full question & answer→Question 743 Marks
Write the expressions for $C_v$ and $C_p$ of a gas in terms of gas constant R and constant $\gamma,$ where $\gamma=\frac{\text{C}_\text{P}}{\text{C}_\nu}$
AnswerWe know that $C_P - C_v = R$ and $\frac{\text{C}_\text{P}}{\text{C}_\nu}=\gamma$ From equation (ii) $\text{C}_\text{P}=\gamma\text{C}_\nu$ and sustituting this value in (i), We have $\gamma\text{C}_\nu-\text{C}_\nu=\text{R}$ $\Rightarrow\text{C}_\nu=\frac{\text{R}}{(\gamma+1)}$
$\therefore\text{C}_\text{P}=\gamma.\text{C}_\nu=\frac{\gamma\text{R}}{(\gamma-1)}$
View full question & answer→Question 753 Marks
A person weighing $60 kg$ takes in $2000\ kcal$ diet in a day. If this energy were to be used in heating the person without any loss, what would be his rise in temperature? Given specific heat of human body is $0.83 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$
AnswerHere, $m = 60kg = 60 \times 10^3g \Delta\text{Q}=2000\text{K cal}=2\times10^6\text{ cal}$ $\Delta\text{Q}=?,\text{C}=0.83\text{cal g}^{-1} {^\circ}\text{C}$ From $\Delta\text{Q}=\text{cm}\Delta\text{T}$ $\Delta\text{T}=\frac{\Delta\text{Q}}{\text{cm}}=\frac{2\times10^6}{0.83\times60\times10^3}$ $=\frac{200}{0.83\times6}=40.16^\circ\text{C}$
View full question & answer→Question 763 Marks
An ideal refrigerator runs between $-23°C$ and $27°C$,
- Find the heat rejected to atmosphere for every joule of work input.
- Also, find heat extracted from cold body.
- Find coefficient of performance of the refrigerator.
AnswerLet heat rejected $Q_1=x$ and $W=1 \mathrm{~J}$ Now, $Q_2=Q_1-W=x-1$ Given, $T_1=273+27=300 \mathrm{~K}, T_2=273-23=250 \mathrm{~K}$
- For an ideal process, $\frac{\text{Q}_2}{\text{Q}_1}=\frac{\text{T}_2}{\text{T}_1}$
$\Rightarrow\frac{\text{x}-1}{\text{x}}=\frac{250}{300}$
$\Rightarrow\text{Q}_1=\text{x}=6\text{J}$
- $Q_2 = 5J$
- Coefficient of performance,
$\beta=\frac{\text{T}_2}{\text{T}_1-\text{T}_2}=\frac{250}{300-250}=5$ View full question & answer→Question 773 Marks
Find the pressure required to compress a gas adiabatically at atmospheric pressure to one fifth of its volume $(\gamma=1.4)$
AnswerHere, $P_1 = 1atm$ Let $V_1 = x c.c.; \text{V}_2=\frac{\text{x}}{5}\text{ c.c.;}$
$\gamma=1.4;\text{ P}=?$ Using the relation, $\text{P}_1\text{V}^\gamma_1=\text{P}_2\text{V}^\gamma_2$
$\therefore\text{P}_2=\text{P}_1\Big(\frac{\text{V}_1}{\text{V}_2}\Big)^\gamma=1\bigg(\frac{\text{x}}{\frac{\text{x}}{5}}\bigg)^{1.4}=(5)^{1.4}$ Taking $\log$ both sides, we get $\log\text{P}_2=1.4\log5$
$=1.4\times0.6990=0.97860$
$\therefore\text{P}_2=9.519\text{ atm}.$
View full question & answer→Question 783 Marks
Differentiate between evaporation and boiling.
AnswerEvaporation is a slow process from the liquid to the gaseous state which takes place at the surface of a liquid and at all temperatures. Boiling is a rapid change of a substance from the liquid to the gaseous state which takes place throughout the mass of the liquid at a definite temperature.
View full question & answer→Question 793 Marks
State Dulong and Petit's law.
AnswerAccording to the Dulong and Petit's law, the specific heat per mole of a chemically pure crystalline solid is approximately $6 cal mol^{-1}K^{-1}$ or roughly $25J mole^{-1} K^{-1}$. This law is obeyed with quite a good approximation for many substances at the room temperature.
View full question & answer→Question 803 Marks
Establish relation between two specific heats of a gas. Which is greater and why?
AnswerRelation between $C_p$ and $C_v$. Suppose one mole of a gas is heated so that its temperature rises by $d T$. Heat supplied $=1 \times C_p \times d T=C_v d T \ldots$...(i) Since the volume is constant, the gas will not perform external work in accordance with the first law of thermodynamics and the heat supplied will be just equal to the increase in the internal energy of the gas. $\mathrm{dU}=\mathrm{C}_{\mathrm{v}} \mathrm{dT}$...(ii) Let the gas be heated at constant pressure to again increase its temperature by dT , and dQ be the amount of heat supplied, therefore, $\therefore \mathrm{dQ}=1 \times \mathrm{C}_p \times d T=C_p d T$...(iii) The heat supplied at a constant pressure increases the temperature by dT hence, increases its internal energy by dU as well as enables the gas to perform work $d W . d W=P V$...(iv) From the first law of thermodynamics, we have, $d Q=d U+d W$ Substituting the values, we get, $C_p d T=C_v d T+$ PDV But PV $=$ RT (For one mole of the gas) $O r$ PdV $=$ RdT $\therefore C_p d T=C_v d T+R d T$ or $C_p-C_v=R \ldots$ (v) This is the relation between two principal specific heats of the gas when $C_p, C_v$ and $R$ are measured in the units of either heat or of work. $C_p>C_v$ because a part of the energy supplied in the adiabatic process goes to increase the volume of the gas and the remaining increases the temperature.
View full question & answer→Question 813 Marks
Calculate the efficiency of Carnot's engine working between steam point and ice point.
AnswerHere, steam point $\text{T}_1=100^\circ\text{C}$ $=100+273=373\text{K}$ and ice point $\text{T}_2=0^\circ\text{C}$ $=0+273=273\text{K}$ $\therefore\eta=1-\frac{\text{T}_2}{\text{T}_1}$ $=1-\frac{273}{373}=\frac{100}{373}$ $\therefore\eta=\frac{100}{373}\times100\%=26.81\%$
View full question & answer→Question 823 Marks
Prove for an adiabatic process:
- $\text{TV}^{\gamma-1}=\text{constant}.$
- $\text{P}^{1-\gamma}\text{T}^\gamma=\text{constant.}$
Answer
- We know $\text{PV}^{\gamma}=$ constant for adiabatic process.
Also, $\text{PV}=\text{nRT}$
$\therefore\text{P}=\frac{\text{nRT}}{\text{V}}$
Replacing P, we have $\frac{\text{nRT}}{\text{RT}}.\text{V}^{\gamma}=$ constant (or) $\text{TV}^{\gamma-1}=$ constant.
- From PV = nRT, we have $\text{V}=\frac{\text{nRT}}{\text{P}}$
Replacing V, we have, $\text{P}\Big(\frac{\text{nRT}}{\text{P}}\Big)^{\gamma}=$ constant.
$\therefore\text{T}^{\gamma}\text{P}^{1-\gamma}=$ constant. View full question & answer→Question 833 Marks
A refrigerator transfers $250J$ heat per second from $-23^\circ C$ to $25^\circ C$. Find the power consumed, assuming no loss of energy.
AnswerHere, $\mathrm{Q}_2=250 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~T}_2=-23^{\circ} \mathrm{C}=-23+273=250 \mathrm{~K}_1=25^{\circ} \mathrm{C}=25+273=298 \mathrm{~K}$
We know, $\beta=\frac{\text{Q}_2}{\text{W}}=\frac{\text{T}_2}{\text{T}_1-\text{T}_2}$
$\therefore\text{W}=\frac{\text{Q}_2(\text{T}_1-\text{T}_2)}{\text{T}_2}$
$\text{W}=\frac{250(298-250)}{250}=\frac{250\times48}{250}$
$\text{W}=48\text{J s}^{-1}$
View full question & answer→