Physics 5 Marks Questions

$\therefore$ Rise in temperature, ΔT = T₂ - T₁

$\therefore$ V₂ = V₁/2

$\therefore\Delta\text{Q} = 0$

$\Delta\text{W} = -22.3\text{ J}$ (Since the work is done on the system)

1. Efficiency of A = efficiency of B

$\eta_\text{A}=\eta_\text{B}$

$\Rightarrow1-\frac{\text{T}}{900}=1-\frac{400}{\text{T}}$

$\Rightarrow\text{T}^2=900\times400$

$\Rightarrow\text{T}=600\text{K}$

2. Let the first engine take Q₁ heat as input at temperature, T1 = 800 K and gives out heat Q₂ at temperature T₀. The second engine receive Q₂ as input and ive is out heat Q₃ at temprature T₃ = 300K to the sink.

Work done by first (A) engine = work done by second (B) engine.

Thus, Q₁ - Q₂ = Q₂ - Q₃

Dividing both sides by Q₁

$1-\frac{\text{Q}_2}{\text{Q}_1}=\frac{\text{Q}_2}{\text{Q}_1}-\frac{\text{Q}_3}{\text{Q}_1}$

$\Rightarrow1-\frac{\text{T}}{\text{T}_1}=\frac{\text{Q}_2}{\text{Q}_1}\Big(1-\frac{\text{Q}_3}{\text{Q}_2}\Big)$

$\Rightarrow1-\frac{\text{T}}{\text{T}_1}=\frac{\text{T}}{\text{T}_1}\Big(1-\frac{\text{T}_3}{\text{T}}\Big)$

$\Rightarrow\frac{\text{T}_1}{\text{T}}-1=1-\frac{\text{T}_3}{\text{T}}$

$\Rightarrow\frac{\text{T}_1}{\text{T}}+\frac{\text{T}_3}{\text{T}}=2$

$\Rightarrow\frac{1}{\text{T}}\Big(\text{T}_1+\text{T}_3)=2$

$\Rightarrow\text{T}=\frac{\text{T}_1+\text{T}_3}{2}$

$\Rightarrow\text{T}=\frac{900+400}{2}=650\text{K}$

1. For an adiabatic process of thermodynamics,

dQ = 0

Since dU = n C_v dT irrespective of the process, from 1st law, we get,

dW = -dU = -n C_V dT

So work done,

W = -nC_V(T_f - T_i)

$=-\frac{\text{C}_\text{V}}{\text{R}}(\text{P}_\text{f}\text{V}_\text{f}-\text{P}_\text{i}\text{V}_\text{i})$

2. For an isothermal process, dT = 0

So, $\text{dW}\int\limits{\text{PdV}}=\int\limits\frac{\text{nRT}}{\text{V}}\text{dV}$

Work done $=\text{W}=\text{nRT}|\log_\text{e}\text{V}|_{\text{V}_\text{i}}^{\text{V}_\text{f}}$

$\text{W}=\text{n}\text{RT}\log_\text{e}\Big(\frac{\text{V}_\text{f}}{\text{V}_\text{i}}\Big)$

1. It is considered that piston is mass less and piston is balanced by atmospheric pressure (P_a). So the initial pressure of system inside the cylinder = P_a,
2. On supply heat Q. Volume of gas increase from V₀ to V₁ and spring stretched also.So increase in volume = V₁ - V₀

If displacement of piston is x then volume increase in cylinder ) 

= Aera of base × height = A × x

A × x = V₁ - V₀ (A = area of cross section of cylinder)

$\therefore\ \text{x}=\frac{\text{V}_1-\text{V}_0}{\text{A}}$

Force exerted by spring $\text{F}_\text{s}=\text{K}_\text{x}=\frac{\text{K}(\text{V}_1-\text{V}_0)}{\text{A}}$

As the piston is of unit area of cross - section $\therefore\ \text{A}=1$

Force due to spring =$\text{K}(\text{V}_1-\text{V}_0)$ on unit area can be say press due to spring = $\text{K}(\text{V}_1-\text{V}_0)$

Final total pressure on gas $\text{P}_\text{f}=\text{P}_\text{a}+\text{K}(\text{V}_1-\text{V}_0)$

3. by 1st law of thermodynamics $\text{dQ}=\text{dU}+\text{dW}$

Now $\text{DQ}=\text{dU}+\text{dW}$

$=\text{c}_\text{v}(\text{T}-\text{T}_0)+\text{P}_\text{a}(\text{V}_1-\text{V}_0)+\frac{1}{2}\text{kx}^2$

$\text{dQ}=\text{C}_\text{V}(\text{T}-\text{T}_0)+\text{P}_\text{a}(\text{V}_1-\text{V}_0)+\frac{1}{2}\text{k}(\text{V}_1-\text{V}_0)^2$

It is required relation.

1. If the temperature of the sink corresponding to $\eta'$ be $\text{T}'_2$ (source at 373k), then,

Thus, $\eta'=1-\frac{\text{T}_2'}{373}$

$\frac{\text{T}_2'}{373}=1-\eta'=1-\frac{873}{1865}=\frac{992}{1865}$

$\text{T}'_2=198.4\text{K}$

$\text{T}_2'-\text{T}_2=198.4\text{K}-273\text{K}=-74.6\text{K}$

2. If the temperature of the source corresponding to $\eta'$ be $\text{T}'_2$ (sinkb at 273K), then

$\eta'=1-\frac{\text{T}'_2}{\text{T}'_1}$ or $\frac{873}{1865}=1-\frac{273}{\text{T}'_1}$

$\frac{273}{\text{T}'_1}=1-\frac{873}{1865}=\frac{992}{1865}$

$\text{T}'_2=513\text{K}$

$\text{T}'_1-\text{T}_1=513\text{K}-373\text{K}$

$=149.25\text{K}=140^\circ\text{C}$

1. Work done during the process from A to B (expansion),

$\text{W}_{\text{AB}}=+\text{area ABKLA}$

= area of $\Delta\text{ABC}+\text{area or rectangle BCLK}$

$\text{W}_{\text{AB}}=\frac{1}{2}\text{BC}\times\text{AC}+\text{KL}\times\text{LC}$

Now, BC = KC = 4 - 1 = 3 litre

= 3 × 10^-3m³

AC = 4 - 2 = 2 Nm^-2

LC = 2 - 0 = 2 Nm^-2

$\therefore\text{W}_{\text{AB}}=\frac{1}{2}\times3\times10^{-3}\times2+3\times10^{-3}\times2$

$\text{W}_{\text{AB}}=9\times10^{-3}\text{J}$

2. Work done during the process from B to C (compression) is,

$\text{W}_{\text{BC}}=-\text{area BCLK}=-\text{KL}\times\text{LC}$

$=-3\times10^{-3}\times2=-6\times10^{-3}\text{J}$

3. Work done during the process from C to A. As there is no change in volume of the gas in this process, therefore,

$\text{W}_{\text{CA}}=0.$

Net work done in the complete cycle,

$\text{W}=\text{W}_{\text{AB}}+\text{W}_\text{BC}+\text{W}_\text{CA}$

$=9\times10^{-3}+(-6\times10^{-3})+0$

$=3\times10^{-3}\text{J}$

1. Adiabatic process: A change in pressure and volume of a gas in which temperature also changes is called an adiabatic change. In such a change, no heat is allowed to enter into or escape from the gas (i.e., there is no exchange of heat between the gas and its surroundings).

Derivation: Consider one gram mole of an ideal gas enclosed in a cylinder with perfectly non-conducting walls, and fitted with a perfectly frictionless, non-conducting piston. P₁ = Let be the initial pressure, v₁ be the volume and T₁ be the temperature of the gas.

Force exerted by the gas on the piston of area of cross-section ‘A’ is given by

F = P × A ...(i)

Where P is the pressure at any instant during expansion.

If we assume that pressure of the gas during an infinitesimally small outward displacement dx of the piston remains constant, then small amount of work done during expansion,

dW = F × dx = (P × A)dx

dW = PdV ...(ii)

Where dV = A(dx) = small increase in the volume of the gas. Total work done by the gas in adiabatic expansion from volume v₁ to v₂ is

$\text{W}=\int\limits^{\text{V}_2}_{\text{V}_1}\text{P dV}\dots\text{(iii)}$

The equation of adiabatic changes is 

PV' = K, a constant ...(iv)

Where, $\gamma=\frac{\text{C}_\text{P}}{\text{C}_\text{V}}$

$=\frac{\text{Specific heat of gas at constant pressure}}{\text{Specific heat of gas at constant volume}}$

For equation (iv)

$\text{P}=\frac{\text{K}}{\text{V}^\gamma}=\text{KV}^{-\gamma},$ put in equation (iii)

$\text{W}=\int\limits^{\text{V}_2}_{\text{v}_1}\text{KV}^{-\gamma}\text{dV}=\text{K}\Big[\frac{\text{V}^{(1-\gamma)}}{1-\gamma}\Big]^{\text{v}_2}_{\text{v}_1}$

$\text{W}=\frac{\text{K}}{1-\gamma}\Big[\text{K}\nu^{(1-\gamma)}_2-\text{K}\nu^{(1-\gamma)}_1\Big]$

$\text{W}=\frac{1}{1-\gamma}\Big[\text{K}\nu^{(1-\gamma)}_2-\text{K}\nu^{(1-\gamma)}_1\Big]\dots\text{(v)}$

From standard gas equation P₁ V₁ = RT, P₂V₂ = RT₂

$\text{W}=\frac{1}{1-\gamma}[\text{RT}_2-\text{RT}_1]$

[Putting above in equation (v)]

$\text{W}=\frac{\text{R}(\text{T}_2-\text{T}_1)}{1-\gamma}$

2. Here, T₂ = 9°C = 9 + 273K = 282K,

T₁ = 36°C = 36 + 273 = 309K

Coeffiecient of performance

$=\frac{\text{T}_2}{\text{T}_1-\text{T}_2}=\frac{282}{309-282}=\frac{282}{27}=10.4$

1. Isothermal process is a change in pressure and volume of a gas without any change in its temperature. In such a change, there is a free exchange of heat between the gas and its surroundings.

For a small change in volume, work done is given by

dW = PdV

We know, PV = nRT

$\therefore\text{P}=\frac{\text{nRT}}{\text{V}}$

For T = constant,

$\text{dW}=\text{nRT}\frac{\text{dV}}{\text{V}}$

Net work done under isothermal condition to change the volume from V_i to V_f is,

$\text{W}=\int\limits^{\text{V}_\text{f}}_{\text{V}_\text{i}}\text{dW}=\text{nRT}\int\limits^{\text{V}_\text{f}}_{\text{V}_\text{i}}\frac{\text{dV}}{\text{V}}$

$=\text{nRT}\big|\log_\text{e}\text{V}\big|^{\text{V}_\text{f}}_{\text{V}_\text{i}}$

$\text{W}=\text{nRT }\log_\text{e}\Big(\frac{\text{V}_\text{f}}{\text{V}_\text{i}}\Big)$

$\therefore\text{W}=2.3026\text{ nRT }\log_\text{10}\Big(\frac{\text{V}_\text{f}}{\text{V}_\text{i}}\Big)$

Where n is the number of moles. If P_f and P_i are the pressures, we can also write,

$\text{W}=2.3026\ \text{nRT }\log_{10}\Big(\frac{\text{P}_\text{i}}{\text{P}_\text{f}}\Big)$

2. Coefficient of performance, $\beta=\frac{\text{T}_2}{\text{T}_1-\text{T}_2}$

$\because\text{T}_1=36 +273=309\text{K}$

and $\text{T}_2=10+273=283\text{K}$

$\therefore\beta=\frac{283}{309-283}=\frac{283}{26}=10.8$

1. Since $\eta=1-\frac{\text{T}_2}{\text{T}_1}$ On increasing T₂, effieciecy drops.
2. Q₁ = 1000J, T₁ = 127 + 273 = 400K, Q₂ = 600J.

1. Effciency $=\eta=\frac{\text{Q}_1-\text{Q}_2}{\text{Q}_1}=1-\frac{\text{T}_2}{\text{T}_1}$

$=\frac{400}{1000}=0.4\text{ or }40\%$

2. Using $\eta=\frac{\text{Q}_1-\text{Q}_2}{\text{Q}_1}=1-\frac{\text{T}_2}{\text{T}_1}$

We get $\frac{400}{1000}=1-\frac{\text{T}_2}{\text{T}_1}$

$\Rightarrow\text{T}_2=400\times\frac{600}{1000}=240\text{K}$

3. Work done = Q₁ - Q₂ = 400J

1. According 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

Limitations of the first law of thermodynamics:

1. The first law does not indicate the direction in which the heat change can occur.
2. The first law does not give any idea about the extent of heat change.
3. The first law of thermodynamics gives no information about the source of heat, i.e. whether it is a hot or cold body.

The relation between two specific heats of a gas, i.e. C_p and C_v is given by C_p - C_v = R where, R is the molar gas constant and is equal to 8.31J mole^-1K^-1.

C_p > C_v because a part of the energy supplied in the isobaric process goes to increase the volume of the gas and the remaining increases the temperature.

2. According to first law of thermodynamics,

dQ = dU + dW

Differentiating w.r.t., time (t)

$\frac{\text{dQ}}{\text{dt}}=\frac{\text{dU}}{\text{dt}}+\frac{\text{dW}}{\text{dt}}$

$100\text{W}=\frac{\text{dU}}{\text{dt}}+75$

$[\because$ Given, system performs work at a rate of 75J/ sec.$]$

$\frac{\text{dU}}{\text{dt}}\rightarrow$ Rate of change of internal energy

$\frac{\text{dU}}{\text{dt}}=100-75=25\text{W}$

Alternate Answer

1. For a small change in volume, work done is given by dW = PdV.

We know, PV = nRT

For T = Constant

$\text{dW}=\text{nRt}\cdot\frac{\text{dV}}{\text{V}}$

$\therefore\text{P}=\frac{\text{nRT}}{\text{V}}$

Net work done under isothermal condition to change the volume from Y_i to V_f is,

$\text{W}=\int\limits^{\text{V}_\text{f}}_{\text{V}_\text{i}}\text{dW}$

$=\text{nRT}\int\limits^{\text{V}_\text{f}}_{\text{V}_\text{i}}\frac{\text{dV}}{\text{V}}=\text{nRT}|\log_\text{e}\text{V}|^{\text{V}_\text{f}}_{\text{V}_\text{i}}$

$=\text{nRT }\log_\text{e}\Big(\frac{\text{V}_\text{f}}{\text{V}_\text{i}}\Big)$

$\therefore\text{W}=2.3026\text{ nRT }\log_{10}\Big(\frac{\text{V}_\text{f}}{\text{V}_\text{i}}\Big)$

2. Work done by a steam energy,

W = 5.4 × 10⁸J, Q₁ = 3.6 × 10⁹J

Efficiency of the engine, $\eta=?,\text{Q}_2=?$

$\eta=\frac{\text{W}}{\text{Q}_1}=\frac{5.4\times10^8\text{J}} {3.6\times10^9\text{J}}$

$=\frac{54}{36}\times10^{-1}=\frac{3}{20}=0.15$ or 15%

Heat wasted per minute,

$\text{Q}_2=\text{Q}_1-\text{W}$

$\text{Q}_2=3.6\times10^9\text{J}-5.4\times10^8\text{J}$

$=36\times10^8\text{J}-5.4\times10^8\text{J}$

$=10^8(36-5.4)\text{J}=30.6\times10^9\text{J}$

1. For path ibf,

Q = 36cal. and dU = U_f - U_i = 30cal

$\therefore\text{W}=\text{Q}-(\text{U}_\text{f}-\text{U}_\text{i})$

$=36-30=6\text{cal}$

2. For path fi.

$\text{W}=-13\text{cal}\text{ dU}=30\text{cal}$

$\therefore\text{Q}=\text{W}+(\text{U}_\text{f}-\text{U}_\text{i})$

3. $\text{U}_\text{i}=10\text{cal}$

$\text{dU}=\text{U}_\text{f}-\text{U}_\text{i}=30$

$\therefore\text{U}_\text{f}=30+\text{U}_\text{i}$

$=30+10=40\text{cal}$

4. For process bf, volume is constant i.e., work done is zero.

$\therefore\text{Q}=\text{dU}=\text{U}_\text{f}-\text{U}_\text{b}$

$=40-22=18\text{cal}$

For path ib,

$\therefore\text{Q}=\text{Q}_{\text{ibf}}-\text{Q}_\text{bf}$

$=36-18=18\text{cal}$

1. Let Q be the heat etracted from the freezing compertment in 10 minutes

$\therefore\text{Q}=\text{Q}_2\times10\text{min}$

$=2250\times10\times60$

$=1350000\text{J}$

$=\frac{135\times10^4}{4.2\times10^3}\text{kcal}$

$=321.4\text{ kcal}$

2. Heat required to convert 1kg of water at 273K into ice,

$\text{Q}'=\text{m}\times\text{L}$

$=1\times80\text{ kcal}$

$=80\times4.2\times10^3\text{J}$

Let Q' be etraction of heat from freezing compartment

$=\frac{80\times4.2\times10^3}{\text{t}}\text{Js}^{-1}$

This rate must be equal to Q₂.

i.e., $2250=\frac{80\times1.2\times10^3}{\text{t}}$

$\therefore\text{t}=\frac{80\times4.2\times10^3}{2250}=149.33\text{s}.$

1. If keeping temperature of colder reservoir fixed the temperature of hot reservoir is changed to $\text{T}_1'$ then

$46.8=\frac{\text{T}_1'-273}{\text{T}'_1}\times100$

$\Rightarrow46.8\text{T}_1'=100$

$\text{T}_1'-27300$

$\Rightarrow53.2\text{T}'_2=27300$ or $\frac{27300}{53.2}=\text{513.2}\text{K}$

$\therefore\text{T}_1'-\text{T}_1=513.2-373=140.2\text{K}$

It means that temprature of hot resevoir be raised by 140.2K

2. If keeping temperature of hot reservoir fixed, the temperature of colder reservoir is changed to $\text{T}_2',$ then

$46.8=\frac{\text{T}_1-\text{T}_2'}{\text{T}_1}\times100$

$=\frac{373-\text{T}'_2}{373}\times100$

$\therefore373\times46.8=373\times100-100\text{T}_2'$

$\Rightarrow100\text{T}_2'=373\times(100-46.8)$

$=373\times53.2$

$\Rightarrow\text{T}'_2=\frac{373-53.2}{100}=198.4\text{K}$

$\therefore\text{T}_2-\text{T}_2'=273-198.4$

$=74.6\text{K}=74.6^\circ\text{C.}$

It means that temparture of colder reservoir be lowered by 74.6°C.

$\therefore\Delta\text{Q} = 0$

$\therefore\ \Delta\text{U}=\Delta\text{Q}-\Delta\text{W}=-(-22.3\text{ J})$

$\Delta\text{U} = +22.3\text{ J}$

$\Delta\text{Q} = 9.35\text{ cal}$

$\therefore\ \Delta\text{W}=\Delta\text{Q}-\Delta\text{U}$

1. : For A → B, dV = 0 

so,$\text{dW}=\int\text{P}.\text{dV}=\int\text{p}\times0=0$

$\text{dW}=0$

By $1^\text{st}$ law of thermodynamics

$\text{dQ}=\text{dU}+\text{dW}=\text{dU}+0$

$\therefore\ \text{dQ} =\text{dU}$

$\big(\text{dQ}=\text{n}\text{C}_\text{V}\text{dT}\big)$

So, $\text{dQ}-1\frac{3}{2}\text{R}\big(\text{T}_\text{B}-\text{T}_\text{A}\big).....(\text{i})$

$\text{dU}=\text{dQ}=\frac{3}{2}\big(\text{R}\text{T}_\text{}B-\text{R}\text{T}_\text{A}\big)=\frac{3}{2}\big(\text{P}_\text{B}\text{V}_\text{B}-\text{P}_\text{A}\text{}V_\text{A}\big)$

$\therefore$  Heat exchange [to system]

$\text{dQ}_1=\text{dU}=\frac{3}{2}\big(\text{P}_\text{s}\text{V}_\text{s}-\text{P}_\text{A}\text{V}_\text{A}\big)$

2. For B to C, $\Delta\text{P}=0\ \text{n}=1$

$\text{dQ}=\text{dU}+\text{dW}=\text{C}_\text{V}\text{(dT)}+\text{P}_\text{S}\text{dV}$

$\text{dQ}_2=\frac{3}{2}\text{R}\big(\text{T}_\text{C}-\text{T}_\text{B}\big)+\text{P}_\text{B}\big(\text{V}_\text{C}-\text{V}_\text{B}\big)$

$=\frac{3}{2}\big(\text{T}_\text{C}\text{R}-\text{RT}_\text{B}\big)+\text{P}_\text{B}\text{V}_\text{C}-\text{P}_\text{B}\text{V}_\text{B}$

$=\frac{3}{2}[\text{P}_\text{C}\text{V}_\text{C}]-\frac{3}{2}[\text{P}_\text{B}\text{V}_\text{B}]-\text{P}_\text{B}\text{V}_\text{B}-\text{P}_\text{B}\text{V}_\text{C}$

$\text{V}_\text{A}=\text{V}_\text{B}\ \text{and}\ \text{P}_\text{B}=\text{P}_\text{C}$

$\therefore\text{dQ}_2=\frac{3}{2}\text{P}_\text{B}\text{V}_\text{C}-\frac{3}{2}\text{P}_\text{B}\text{V}_\text{A}-\text{P}_\text{B}\text{V}_\text{A}+\text{P}_\text{B}\text{V}_\text{C}$

$=\frac{5}{2}\text{P}_\text{B}\text{V}_\text{C}-\frac{5}{2}\text{P}_\text{B}\text{V}_\text{A}$

$\text{dQ}_2=\frac{5}{2}\text{P}_\text{B}[\text{V}_\text{C}-\text{V}_\text{A}]$

3.  For diagram C → B,  adiabatic change

$\text{dQ}_3=0$  (No exchange of heat )

4.  For diagram D  →  A, $\Delta\text{P}=0$ Compression of gas from volume V_D to V_A  pressure hence heat exchange similar to part (b) i.e. Heat exchange$\text{dQ}_3=\frac{5}{2}\text{P}_\text{A}\big(\text{V}_\text{A}-\text{V}_\text{D}\big)$

(a) A to B

(b) C to D

(c) $\text{W}_\text{AB}=\int\limits^\text{B}_\text{A}\text{pdV}=0;\text{W}_\text{CD}=0.$

Simillarly. $\text{W}_\text{BC}=\Big[\int\limits^\text{C}_\text{B}\text{pdV}=\text{k}\int\limits^\text{C}_\text{B}\frac{\text{dV}}{\text{V}^\text{r}}=\text{k}\frac{\text{V}^\text{-r+1}}{-\text{R}+1}\Big]^{\text{V}_{\text{C}}}_{\text{V}_\text{B}}$

$= \frac{1}{1-\gamma}(\text{P}_\text{c}\text{V}_\text{c}-\text{P}_\text{B}\text{V}_\text{B})$

Simillarly, $\text{W}_\text{DA}=\frac{1}{1-\gamma}(\text{P}_\text{A}\text{V}_\text{A}-\text{P}_\text{D}\text{V}_\text{D})$

Now $\text{P}_\text{C}=\text{P}_\text{B}\Big(\frac{\text{V}_\text{B}}{\text{V}_\text{C}}\Big)^\gamma=2^{-\gamma}\text{P}_\text{B}$

Simillarly, $\text{P}_\text{D}=\text{P}_\text{A}2^{-\gamma}$

Total work done $=\text{W}_\text{BC}+\text{W}_\text{DA}$

$=\frac{1}{1-\gamma}\big[\text{P}_\text{B}\text{V}_\text{B}\big(2^{-\gamma+1}-1\big)-\text{P}_\text{A}\text{V}_\text{A}\big(2^{-\gamma+1}-1\big)\big]$

$=\frac{1}{1-\gamma}\big(2^{1-\gamma}-1\big)\big(\text{P}_\text{B}-\text{P}_\text{A}\big)\text{V}_\text{A}$

$=\frac{3}{2}\big(1-\Big(\frac{1}{2}\Big)^\frac{2}{3}\big)\big(\text{P}_\text{B}-\text{P}_\text{A}\big)\text{V}_\text{A}$

4. Heat supplied during process A, B 

$\text{d}\text{Q}_\text{AB}=\text{d}\text{U}_\text{AB}$

$\text{Q}_\text{AB}=\frac{3}{2}\text{n}\text{R}\big(\text{T}_\text{B}-\text{T}_\text{A}\big)\text{V}_\text{A}$

$\text{Efficiency}=\frac{\text{Net Work done }}{\text{Heat Supplied}}=\Big[1-\Big(\frac{1}{2}\Big)^\frac{2}{3}\Big]$