Thermodynamics MCQ Questions & Answers in Heat and Thermodynamics | Physics

Under adiabatic change
$$\eqalign{ & \frac{{{T_2}}}{{\;{T_1}}} = {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{\frac{{1 - \gamma }}{\gamma }}}\,\,{\text{or}}\,\,{T_2} = {T_1}{\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{\frac{{1 - \gamma }}{\gamma }}} \cr & \therefore {T_2} = 300{\left( {\frac{4}{1}} \right)^{\frac{{1 - \left( {\frac{7}{5}} \right)}}{{\left( {\frac{7}{5}} \right)}}}}; \cr & \gamma = 1.4 = \frac{7}{5}\,{\text{for}}\,{\text{air}} \cr & {\text{or}}\,\,{T_2} = 300{\left( 4 \right)^{ - \frac{2}{7}}} \cr} $$

The efficiency $$\left( \eta \right)$$ of a Carnot engine and the co - efficient of performance $$\left( \beta \right)$$ of a refrigerator are related as
$$\eqalign{ & \beta = \frac{{1 - \eta }}{\eta } \cr & {\text{Here, }}\eta = \frac{1}{{10}} \cr & \therefore \,\,\beta = \frac{{1 - \frac{1}{{10}}}}{{\left( {\frac{1}{{10}}} \right)}} \cr & = 9. \cr} $$
Also, Co - efficient of performance $$\left( \beta \right)$$  is given by $$\beta = \frac{{{Q_2}}}{W},$$
where $${{Q_2}}$$ is the energy absorbed from the reservoir.
$$\eqalign{ & {\text{or, }}9 = \frac{{{Q_2}}}{{10}} \cr & \therefore \,\,{Q_2} = 90\,J. \cr} $$

The heat is supplied at constant pressure. Therefore,
$$\eqalign{ & Q = n{C_p}\,\Delta \,t \cr & = 2\left[ {\frac{5}{2}R} \right] \times \Delta\, t \cr & = 2 \times \frac{5}{2} \times 8.31 \times 5 \cr & = 208\,J \cr & \left( {\because \,\,{C_p} = \frac{5}{2}\,R{\text{ for mono - atomic gas}}} \right) \cr} $$

Heat given to system $$ = {\left( {n{C_{V\Delta }}\,T} \right)_{A \to B}} + {\left( {n{C_P}\,\Delta T} \right)_{B \to C}}$$
$$\eqalign{ & = {\left[ {\frac{3}{2}\left( {nR\,\Delta T} \right)} \right]_{A \to B}} + {\left[ {\frac{5}{2}\left( {nR\,\Delta T} \right)} \right]_{B \to C}} \cr & = {\left[ {\frac{3}{2} \times {V_0}\,\Delta P} \right]_{A \to B}} + \left[ {\frac{5}{2} \times 2{P_0} \times {V_0}} \right] \cr & = \frac{{13}}{2}{P_0}{V_0} \cr & {\text{and }}{W_0} = {P_0}{V_0} \cr & \eta = \frac{{{\text{Work}}}}{{{\text{heat given}}}} \cr & = \frac{{{P_0}{V_0}}}{{\frac{{13}}{2}{P_0}{V_0}}} \times 100 \cr & = 15.4\% \cr} $$

Key Concept
Coefficient of performance $$\left( \beta \right)$$ of a refrigerator is defined as the ratio of quantity of heat removed per cycle $$\left( {{Q_2}} \right)$$ to the work done on the working substance per cycle to remove this heat.
Given, coefficient of performance of a refrigerator, $$\beta = 5$$
Temperature of surface, i.e. inside freezer,
$${T_2} = - {20^ \circ }C = - 20 + 273 = 253\,K$$
Temperature of surrounding, i.e. heat rejected outside $${T_1} = ?$$
$$\eqalign{ & {\text{So,}}\,\,\beta = \frac{{{T_2}}}{{{T_1} - {T_2}}} \Rightarrow 5 = \frac{{253}}{{{T_1} - 253}} \cr & \Rightarrow 5{T_1} - 1265 = 253 \cr & \Rightarrow 5{T_1} = 1518 \cr & {T_1} = \frac{{1518}}{5} = 303.6\,K \cr & {T_1} = 303.6 - 273 = {31^ \circ }C \cr} $$

$$\eqalign{ & W = \frac{{nR\Delta T}}{{1 - \gamma }} \cr & \Rightarrow \,\, - 146000 = \frac{{1000 \times 8.3 \times 7}}{{1 - \gamma }} \cr & {\text{or }}1 - \gamma = - \frac{{58.1}}{{146}} \cr & \Rightarrow \,\,\gamma = 1 + \frac{{58.1}}{{146}} \cr & = 1.4 \cr} $$
Hence the gas is diatomic.

Efficiency of the Carnot engine is given by
$$\eta = 1 - \frac{{{T_2}}}{{{T_1}}}\,......\left( {\text{i}} \right)$$
where, $${{T_1}} =$$  temperature of source
$${{T_2}} =$$  temperature of sink
Given, $$\eta = 50\% = 0.5,\,{T_2} = 500\,K$$
Substituting in Eq. (i), we have
$$\eqalign{ & 0.5 = 1 - \frac{{500}}{{{T_1}}}\,\,{\text{or}}\,\,\frac{{500}}{{{T_1}}} = 0.5 \cr & \therefore {T_1} = \frac{{500}}{{0.5}} = 1000\,K \cr} $$
Now, the temperature of sink is changed to $${{T'}_2}$$ and the efficiency becomes $$60\% $$  i.e., 0.6.
Using Eq. (i), we get
$$\eqalign{ & 0.6 = 1 - \frac{{{{T'}_2}}}{{1000}} \cr & {\text{or}}\,\frac{{{{T'}_2}}}{{1000}} = 1 - 0.6 = 0.4\,\,{\text{or}}\,\,{{T'}_2} = 0.4 \times 100 = 4000\,K \cr} $$
NOTE
Carnot engine is not a practical engine because many ideal situations have been assumed while designing this engine which cannot be obtained practically.

According to the first law of thermodynamics, when some quantity of heat $$\left( {dQ} \right)$$  is supplied to a system capable of doing external work, then the quantity of heat absorbed by the system $$\left( {dQ} \right)$$  is equal to the sum of the increase in the internal energy of the system $$\left( {dU} \right)$$  due to rise in temperature and the external work done by the system $$\left( {dQ} \right)$$  in expansion,
i.e. $$dQ = dU + dW$$
This law, which is basically the law of conservation of energy applies to every process in nature.

According to adiabatic process the relation between temperature and volume is given by $$T{V^{\gamma - 1}} = {\text{constant}}$$
So, for two different cases
$$\therefore {T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}\,......\left( {\text{i}} \right)$$
Given, initial temperature $${T_1} = {18^ \circ }C = 291\,K$$
Let initial volume $${V_1} = V$$
and final volume $${V_2} = \left( {\frac{1}{8}} \right)V$$
Putting these values in Eq. (i)
$$\eqalign{ & {T_2} = 291{\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}} = 291{\left( {\frac{V}{{\left( {\frac{1}{8}} \right)V}}} \right)^{\frac{7}{5} - 1}}\,\,\left( {\gamma = \frac{7}{5}{\text{for}}\,{\text{diatomic}}\,{\text{gas}}} \right) \cr & = 291 \times 2.297 = 668.4\,K \cr} $$

Isobaric compression is represented by curve $$AO$$
Work done = area under $$AD$$
$$ = 2 \times {10^2} \times \left( {3 - 1} \right) = 4 \times {10^2} = 400\,J.$$