Thermodynamics MCQ Questions & Answers in Heat and Thermodynamics | Physics

$$\eqalign{ & P = \frac{{dQ}}{{dt}} \cr & = \frac{d}{{dt}}\left( {mc} \right)T \cr & = \left( {mc} \right)\frac{{dT}}{{dt}} \cr & = \left( {mc} \right)\frac{d}{{dt}}\left[ {{T_0}\left( {1 + \beta {t^{\frac{1}{4}}}} \right)} \right] \cr & P = \left( {mc} \right){T_0} \cr} $$
$$ = \frac{{\beta {t^{ - \frac{3}{4}}}}}{4}$$   where $$(mc)$$  is the heat capacity
$$\eqalign{ & \therefore \,\,\left( {mc} \right) = \frac{{4P{t^{\frac{3}{4}}}}}{{{T_0}\beta }}\,\,\,\,\,.....\left( {\text{i}} \right) \cr & {\text{But }}{t^{\frac{1}{4}}} = \frac{{T\left( t \right) - {T_0}}}{{\beta {T_0}}} \cr & \therefore \,\,{t^{\frac{3}{4}}} = \frac{{{{\left[ {T\left( t \right) - {T_0}} \right]}^3}}}{{{\beta ^3}T_0^3}}\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
From (i) & (ii),
$$\eqalign{ & \left( {mc} \right) = \frac{{4P}}{{{T_0}\beta }}\frac{{{{\left[ {T\left( t \right) - {T_0}} \right]}^3}}}{{{\beta ^3}T_0^3}} \cr & = \frac{{4P{{\left[ {T\left( t \right) - {T_0}} \right]}^3}}}{{{\beta ^4}T_0^4}} \cr} $$

Using $${V_{{\text{rms}}}} = \sqrt {\frac{{\gamma RT}}{M}} $$
$$\eqalign{ & {\text{we have,}} \cr & \frac{{{V_{{\text{rms}}}}\left( {He} \right)}}{{{V_{{\text{rms}}}}\left( {Ar} \right)}} = \sqrt {\frac{{{M_{Ar}}}}{{{M_{He}}}}} \cr & = \sqrt {\frac{{40}}{4}} \cr & = 3.16 \cr} $$

The coefficient of performance of a refrigerator is given by
$$\alpha = \frac{{{Q_2}}}{W} = \frac{{{Q_2}}}{{{Q_1} - {Q_2}}}$$
Substituting the given values, we get
$$\eqalign{ & \frac{1}{3} = \frac{{{Q_2}}}{{200 - {Q_2}}} \cr & \Rightarrow 200 - {Q_2} = 3{Q_2} \Rightarrow 4{Q_2} = 200 \cr & {\text{or}}\,\,{Q_2} = \frac{{200}}{4}J = 50\,J \cr & \therefore W = {Q_1} - {Q_2} = 200\,J - 50\,J = 150\,J \cr} $$

From $$C$$ to $$D,V$$  is constant. So process is isochoric. From $$D$$ to $$A,$$ the curve represents constant temperature. So the process is isothermal.
From $$A$$ to $$B,$$ pressure is constant . So, the process is isobaric.
$$BC$$ represents constant entropy.

Change in entropy is given by
$$\eqalign{ & dS = \frac{{dQ}}{T}\,\,{\text{or}}\,\,\Delta S = \frac{{\Delta Q}}{T} = \frac{{m{L_f}}}{{273}} \cr & \Delta S = \frac{{1000 \times 80}}{{273}} = 293\,cal/K. \cr} $$

Efficiency of heat engine is, $$\eta = 1 - \frac{{{T_2}}}{{{T_1}}}\,\,{\text{or}}\,\,\eta = \frac{{{T_1} - {T_2}}}{{{T_1}}}$$
$${{T_2}} =$$  temperature of sink
$${{T_1}} =$$  temperature of source
Given, $${T_1} = 273 + 127 = 400\,K$$
$${T_2} = 273 + 27 = 300\,K$$
$$\therefore \eta = \frac{{400 - 300}}{{400}} = \frac{{100}}{{400}} = 0.25 = 25\% $$
Hence, $$26\% $$  efficiency is impossible for a given heat engine.

$$\eqalign{ & {Q_{AB}} = \Delta {U_{AB}} + {W_{AB}} \cr & {W_{AB}} = 0 \cr & \Delta {U_{AB}} = \frac{f}{2}nR\Delta T \Rightarrow \frac{f}{2}\left( {\Delta PV} \right) \cr & \Delta {U_{AB}} = \frac{5}{2}\left( {\Delta PV} \right) \Rightarrow {Q_{AB}} = 2.5{P_0}{V_0} \cr} $$
Process $$BC$$ :
$$\eqalign{ & {Q_{BC}} = \Delta {U_{BC}} + {W_{BC}} = 0 + 2{P_0}{V_0}\ln 2 = 1.4{P_0}{V_0} \cr & {Q_{{\text{net}}}} = {Q_{AB}} + {Q_{BC}} = 3.9{P_0}{V_0} \cr} $$

Concept
Apply first law of thermodynamics to calculate the required work done.
From first law of thermodynamics $$\Delta Q = \Delta U + \Delta W$$
where, $$\Delta Q =$$  heat given
$$\Delta U =$$  change in intemal energy
$$\Delta W =$$  work done
Here, $$\Delta Q = 110\,J$$
$$\eqalign{ & \Delta U = 40\,J \cr & \therefore \Delta W = \Delta Q - \Delta U = 110 - 40 = 70\,J \cr} $$

$$\eqalign{ & P\, \propto \,{T^3} \cr & \Rightarrow \,\,P{T^{ - 3}} = {\text{constant }}.....\left( {\text{i}} \right) \cr} $$
But for an adiabatic process, the pressure temperature relationship is given by
$$\eqalign{ & {P^{1 - \gamma }}{T^\gamma } = {\text{constant}} \cr & \Rightarrow {\text{ }}P{T^{\frac{\gamma }{{1 - \gamma }}}} = {\text{constant }}.....\left( {{\text{ii}}} \right) \cr} $$
From (i) and (ii)
$$\eqalign{ & \frac{\gamma }{{1 - \gamma }} = - 3 \cr & \Rightarrow \,\,\gamma = - 3 + 3\gamma \cr & \Rightarrow \,\,\gamma = \frac{3}{2} \cr} $$

$$\eqalign{ & \frac{{{n_1} + {n_2}}}{{\gamma - 1}} = \frac{{{n_1}}}{{{\gamma _1} - 1}} + \frac{{{n_2}}}{{{\gamma _2} - 1}} \cr & \Rightarrow \,\,\frac{{1 + 1}}{{\gamma - 1}} = \frac{1}{{\frac{5}{3} - 1}} + \frac{1}{{\frac{7}{5} - 1}} \cr & \Rightarrow \,\,\gamma = \frac{3}{2} \cr} $$