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.
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