Thermodynamics MCQ Questions & Answers in Heat and Thermodynamics | Physics

Internal energy and volume depend upon states.

$$PV = {\text{constant}}{\text{.}}$$    Differentiating,
$$\frac{{PdV}}{{dP}} = - V;\beta = - \left( {\frac{1}{V}} \right)\left( {\frac{{dV}}{{dP}}} \right) = \left( {\frac{1}{P}} \right) \Rightarrow \beta \times P = 1$$
$$\therefore $$ Graph between $$\beta $$ and $$P$$ will be a rectangular hyperbola.

An isothermal process is a constant temperature process. In this process, $$T = $$  constant or $$\Delta T = 0.$$
$$\eqalign{ & \therefore \Delta Q = \Delta U + \Delta W \cr & \Rightarrow \Delta Q = \Delta W\,\,\left( {\Delta U = 0} \right) \cr & \Delta U = n{C_V}\Delta T = 0 \cr} $$
An adiabatic process is defined as one with no heat transfer into or out of a system. Therefore, $$Q = 0.$$  From the first law of thermodynamics.
$$\eqalign{ & \Delta Q = \Delta U + \Delta W \cr & {\text{or}}\,\,\Delta U = - W\,\,\left[ {\Delta Q = 0} \right] \cr} $$

According to Mayer's relationship $${C_P} - {C_V} = R$$
$$\therefore \,\,\frac{{{C_P}}}{M} - \frac{{{C_V}}}{M} = \frac{R}{M}\,\,\,\,{\text{Here }}M = 28.$$

First law is applicable to a cyclic process. Concept of entropy is introduced by the second law.

In an adiabatic process
$$p =$$  pressure
$$V =$$  volume
$$\gamma = $$  atomicity of gas
$$p{V^\gamma } = {\text{constant}}\,......\left( {\text{i}} \right)$$
Now from ideal gas equation,
$$\eqalign{ & pV = RT\,\,\,\,\left( {{\text{for}}\,{\text{one}}\,{\text{mole}}} \right) \cr & {\text{or}}\,\,p = \frac{{RT}}{V}\,......\left( {{\text{ii}}} \right)\,\left( {R = {\text{gas}}\,{\text{constant}}} \right) \cr} $$
From Eqs. (i) and (ii), we have
$$\eqalign{ & \left( {\frac{{RT}}{V}} \right){V^\gamma } = {\text{constant}} \cr & T{V^{\gamma - 1}} = {\text{constant}}\,......\left( {{\text{iii}}} \right) \cr} $$
So for two different cases of temperature and volume
$$\eqalign{ & {\text{So,}}\,\,{T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1} \cr & {\text{or}}\,\,\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}}\,......\left( {{\text{iv}}} \right) \cr & {\text{Given,}}\,\,{T_1} = {27^ \circ }C = 27 + 273 = 300\,K \cr & {\text{Given,}}\,\,\frac{{{V_2}}}{{{V_1}}} = \frac{8}{{27}},\gamma = \frac{5}{3} \cr} $$
Substituting in Eq. (i), we get
$$\eqalign{ & \frac{{{T_2}}}{{300}} = {\left( {\frac{{27}}{8}} \right)^{\frac{5}{3} - 1}} \cr & {\text{or}}\,\,\frac{{{T_2}}}{{300}} = {\left[ {{{\left( {\frac{3}{2}} \right)}^3}} \right]^{\frac{2}{3}}}{\text{ or }}\frac{{{T_2}}}{{300}} = {\left( {\frac{3}{2}} \right)^2} = \frac{9}{4} \cr & \therefore {T_2} = \frac{9}{4} \times 300 = 675\;K = {402^ \circ }C \cr} $$
Thus, rise in temperature $$ = {T_2} - {T_1} = 402 - 27 = {375^ \circ }C$$

Since, initial and final points are same
So, $$\Delta {U_{A \to B \to C}} = \Delta {U_{A \to C}}\,......\left( {\text{i}} \right)$$
Also $$A \to B$$  is isochoric process
So $$d{W_{A \to B}} = 0$$
and $$dQ = dU + dW$$
So, $$d{Q_{A \to B}} = d{U_{A \to B}} = 400\,J$$

Next $$B \to C$$  is isobaric process
$$\eqalign{ & {\text{So,}}\,\,\,d{Q_{B \to C}} = d{U_{B \to C}} + d{W_{B \to C}} \cr & = d{U_{B \to C}} + p\Delta {V_{B \to C}} \cr & \Rightarrow 100 = d{U_{B \to C}} + 6 \times {10^4}\left( {2 \times {{10}^{ - 3}}} \right) \cr & \Rightarrow d{U_{B \to C}} = 100 - 120 = - 20\,J \cr} $$
From Eq. (i),
$$\eqalign{ & \because \Delta {U_{A \to B \to C}} = \Delta {U_{A \to C}} \cr & \Rightarrow \Delta {U_{A \to B}} + \Delta {U_{B \to C}} = d{Q_{A \to C}} - d{W_{A \to C}} \cr & \Rightarrow 400 + ( - 20) = d{Q_{A \to C}} - \left( {p\Delta {V_A} + } \right.{\text{Area}}\,{\text{of}}\left. {\Delta ABC} \right) \cr & \Rightarrow d{Q_{A \to C}} = 380 + \left( {2 \times {{10}^4} \times 2 \times {{10}^{ - 3}} + \frac{1}{2} \times 2 \times {{10}^{ - 3}} \times 4 \times {{10}^4}} \right) \cr & = 380 + \left( {40 + 40} \right) \cr & d{Q_{A \to C}} = 460\,J \cr} $$

$$0.4 = 1 - \frac{{{T_2}}}{{500}}\,{\text{and }}0.6 = 1 - \frac{{{T_2}}}{{{T_1}}}$$
on solving we get $${T_2} = 750\,K$$

Efficiency of Carnot engine $$\left( {{\eta _1}} \right) = 40\% = 0.4$$
Initial intake temperature $$\left( {{T_1}} \right) = 500\,K$$
and new efficiency $$\left( {{\eta _2}} \right) = 50\% = 0.5$$
Efficiency of Carnot engine is given by
$$\eta = 1 - \frac{{{T_2}}}{{{T_1}}}\,......\left( {\text{i}} \right)$$
where, $${{T_1}}$$ is temperature of source, $${{T_2}}$$ is temperature of sink.
From Eq. (i), $$\frac{{{T_2}}}{{{T_1}}} = 1 - \eta $$
Therefore, in first case $$\frac{{{T_2}}}{{500}} = 1 - 0.4 = 0.6$$
$$ \Rightarrow {T_2} = 0.6 \times 500 = 300\,K$$
and in second case $$\frac{{300}}{{{T_1}}} = 1 - 0.5 = 0.5$$
$$ \Rightarrow {T_1} = \frac{{300}}{{0.5}} = 600\,K$$

For tiny glass tube,
$$\eqalign{ & {P_1}{V_1} = {P_2}{V_2} \cr & \therefore {P_2} = \frac{{{P_1}{V_1}}}{{{V_2}}} = \frac{{4.5 \times 0.50}}{{500}} = 0.0045\,atm. \cr & {\text{Thus}}\,P = 1\,atm + 0.0045\,atm \cr & = 1.0045\,atm = 76.34\,cm\,{\text{of}}\,Hg. \cr} $$