Thermodynamics MCQ Questions & Answers in Heat and Thermodynamics | Physics

Given, ideal gas is compressed to half its initial volume i.e.
$${V_0} = \frac{V}{2}$$

The isochoric process is one in which volume is kept constant, meaning that work done by the system will be zero. $${W_{{\text{isochoric}}}} = 0$$
As we know, work done on the gas = Area under curve, i.e.
$${W_{{\text{adiabatic}}}} > {W_{{\text{isothermal}}}} > {W_{{\text{isobaric}}}}$$

Here $$Q = 0$$   and $$W = 0.$$  Therefore from first law of thermodynamics $$\Delta U = Q + W = 0$$
$$\therefore $$ Internal energy of the system with partition = Internal energy of the system without partition.
$$\eqalign{ & {n_1}{C_v}{T_1} + {n_2}{C_v}{T_2} = \left( {{n_1} + {n_2}} \right){C_v}T \cr & \therefore T = \frac{{{n_1}{T_1} + {n_2}{T_2}}}{{{n_1} + {n_2}}} \cr & {\text{But}}\,{n_1} = \frac{{{P_1}{V_1}}}{{R{T_1}}}\,{\text{and}}\,{n_2} = \frac{{{P_2}{V_2}}}{{R{T_2}}} \cr & \therefore T = \frac{{\frac{{{P_1}{V_1}}}{{R{T_1}}} \times {T_1} + \frac{{{P_2}{V_2}}}{{R{T_2}}} \times {T_2}}}{{\frac{{{P_1}{V_1}}}{{R{T_1}}} + \frac{{{P_2}{V_2}}}{{R{T_2}}}}} = \frac{{{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)}}{{{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}} \cr} $$

$$\eqalign{ & {Q_a} = \Delta U + 0 = \Delta U\,\,{\text{and}}\,\,{Q_b} = \Delta U + P\Delta V \cr & {\text{As}}\,\,{Q_b} > {Q_a} \cr} $$
∴ Change in entropy is greater in case (b).

The adiabatic relation between $$p$$ and $$V$$ for a perfect gas is $$p{V^\gamma } = k\,......\left( {\text{i}} \right)$$
From standard gas equation $$pV = RT\,\,{\text{or}}\,\,V = \frac{{RT}}{p}$$
Putting value of $$V$$ in Eq. (i)
$$\eqalign{ & p{\left( {\frac{{RT}}{p}} \right)^\gamma } = k \cr & {\text{or}}\,\,{p^{1 - \gamma }}{T^\gamma } = \frac{k}{{{R^\gamma }}} = {\text{another constant}} \cr & {\text{i}}{\text{.e}}{\text{.,}}\,\,{p^{1 - \gamma }}{T^\gamma } = {\text{constant}} \cr} $$

$$\frac{{{\text{Slope}}\,{\text{of}}\,{\text{adiabatic}}\,{\text{curve}}}}{{{\text{Slope}}\,{\text{of}}\,{\text{isothermal}}\,{\text{curve}}}} = \frac{{{{\left( {\frac{{dP}}{{dV}}} \right)}_{{\text{adi}}}}}}{{{{\left( {\frac{{dP}}{{dV}}} \right)}_{{\text{iso}}}}}} = + \gamma $$
So slope to adiabatic curve is $$\gamma \left( { = \frac{{{C_p}}}{{{C_v}}}} \right)$$   times of isothermal curve, as clear also from figure.

According to question,
$$p \propto {T^3}\,......\left( {\text{i}} \right)\left( {_{T = {\text{temperature}}}^{p\, = \,{\text{pressure}}}} \right)$$
and we know that
$$pV = nRT\,\,{\text{and}}\,\,pV \propto T\,......\left( {{\text{ii}}} \right)$$
So, putting Eq. (ii) in (i),
$$\eqalign{ & p \propto {\left( {pV} \right)^3} \cr & \Rightarrow {p^2}{V^3} = {\text{constant}} \cr & \Rightarrow p{V^{\frac{3}{2}}} = {\text{constant}}\,......\left( {{\text{iii}}} \right) \cr} $$
$$ \Rightarrow $$ Comparing Eq. (iii) with $$pV\gamma = {\text{constant}}{\text{.}}$$
We have $$\gamma = \frac{3}{2}.$$

The efficiency of cycle is $$\eta = 1 - \frac{{{T_2}}}{{{T_1}}}$$
For adiabatic process $$T{V^{\gamma - 1}} = {\text{constant}}$$
For diatomic gas $$\gamma = \frac{7}{5}$$
$$\eqalign{ & {T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1} \cr & {T_1} = {T_2}{\left( {\frac{{{V_2}}}{{{V_1}}}} \right)^{\gamma - 1}} \cr & {T_1} = {T_2}{\left( {32} \right)^{\frac{7}{5} - 1}} = {T_2}{\left( {{2^5}} \right)^{\frac{2}{5}}} = {T_2} \times 4 \cr & {T_1} = 4{T_2} \cr & \therefore \eta = \left( {1 - \frac{1}{4}} \right) = \frac{3}{4} = 0.75 \cr} $$

According to the Carnot cycle in heat engine $$\frac{{{Q_2}}}{{{Q_1}}} = \frac{{{T_2}}}{{{T_1}}}$$
Given, heat absorbed, $${Q_1} = 6 \times {10^4}cal,$$
Temperature of source, $${T_1} = 227 + 273 = 500\,K$$
Temperature of sink, $${T_2} = 127 + 273 = 400\,K$$
$$\therefore \frac{{{Q_2}}}{{6 \times {{10}^4}}} = \frac{{400}}{{500}}$$
⇒ Heat rejected, $${Q_2} = \frac{4}{5} \times 6 \times {10^4} = 4.8 \times {10^4}cal$$
Now, heat converted to work
$$\eqalign{ & W = {Q_1} - {Q_2} = 6.0 \times {10^4} - 4.8 \times {10^4} \cr & = 1.2 \times {10^4}cal \cr} $$

$$\eqalign{ & {P_1}{V_1}^\gamma = {P_2}{V_2}^\gamma \,\left( {{\text{Adiabatic change}}} \right) \cr & {P_2} = {P_1}{\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^\gamma } = {P_1}{\left( {\frac{{{V_1}}}{{\frac{{{V_1}}}{3}}}} \right)^\gamma } = {P_2}{\left( 3 \right)^\gamma } \cr} $$

$$\eqalign{ & dW = P\Delta V = 1.01 \times {10^5}\left[ {1671 - 1} \right] \times {10^{ - 6}}\,Joule \cr & = \frac{{1.01 \times 167}}{{4.2}}cal. = 40\,cal.\,{\text{nearly}} \cr & \Delta Q = mL = 1 \times 540, \cr & \Delta Q = \Delta W + \Delta Q \cr & {\text{or}}\,\,\Delta U = 540 - 40 = 500\,cal. \cr} $$