Thermodynamics MCQ Questions & Answers in Heat and Thermodynamics | Physics

$$\eqalign{ & {\text{As,}}\,\,{Q_1} + W = {Q_2} \cr & {\text{Given,}}\,\,\eta = \frac{1}{{10}} \cr & {\text{Now,using}}\,\,\eta = 1 - \frac{{{T_1}}}{{{T_2}}} \cr & {\text{So,}}\,\,\frac{1}{{10}} = 1 - \frac{{{T_1}}}{{{T_2}}} \Rightarrow \frac{{{T_1}}}{{{T_2}}} = \frac{9}{{10}} \cr & {\text{Now}}\,\,\frac{{{Q_1}}}{{{Q_2}}} = \frac{{{T_1}}}{{{T_2}}} \Rightarrow \frac{{{Q_1}}}{{{Q_1} + W}} = \frac{9}{{10}} \cr & \Rightarrow 10{Q_1} = 9{Q_1} + 9W \cr & \Rightarrow {Q_1} = 9W = 9 \times 10 = 90\;J \cr} $$

$$\eqalign{ & \frac{{{n_1} + {n_2}}}{{r - 1}} = \frac{{{n_1}}}{{{r_1} - 1}} + \frac{{{n_2}}}{{{r_2} - 1}} \cr & \frac{{\frac{{16}}{4} + \frac{{16}}{{32}}}}{{r - 1}} = \frac{{\frac{{16}}{4}}}{{\frac{5}{3} - 1}} + \frac{{\frac{{16}}{{32}}}}{{1.4 - 1}} \cr & \therefore \,\,\gamma = 1.62 \cr} $$

Work done is equal to area under the curve on $$PV$$  diagram.

For a refrigerator, we know that $$\frac{{{Q_1}}}{W} = \frac{{{Q_1}}}{{{Q_1} - {Q_2}}} = \frac{{{T_1}}}{{{T_1} - {T_2}}}$$
where,
$${{Q_1}} =$$  amount of heat delivered to the room
$$W =$$  electrical energy consumed
$${{T_1}} =$$  room temperature $$= {{t_1} + 273}$$
$${{T_2}} =$$  temperature of sink $$= {{t_2} + 273}$$
$$\therefore \frac{{{Q_1}}}{1} = \frac{{{t_1} + 273}}{{{t_1} + 273 - \left( {{t_2} + 273} \right)}} \Rightarrow {Q_1} = \frac{{{t_1} + 273}}{{{t_1} - {t_2}}}$$

Efficiency of engine
$$\eqalign{ & \frac{1}{6} = 1 - \frac{{{T_2}}}{{{T_1}}}\,{\text{and }}{\eta _2} = 1 - \frac{{{T_2} - 62}}{{{T_1}}} = \frac{1}{3} \cr & \therefore \,\,{T_1} = 372\,K\,\,{\text{and }}{T_2} = \frac{5}{6} \times 372 = 310\,K \cr} $$

Heat lost by $$He$$ = Heat gained by $${N_2}$$
$$\eqalign{ & {n_1}{C_{{v_1}}}\Delta {T_1} = {n_2}{C_{{v_2}}}\Delta {T_2} \cr & \frac{3}{2}R\left[ {\frac{7}{3}{T_0} - {T_f}} \right] = \frac{5}{2}R\left[ {{T_f} - {T_0}} \right] \cr & \Rightarrow \,\,{T_f} = \frac{3}{2}{T_0} \cr} $$

$$\eqalign{ & {T_1} = T,W = 6R\,joule,\gamma = \frac{5}{3} \cr & W = \frac{{{P_1}{V_1} - {P_2}{V_2}}}{{\gamma - 1}} = \frac{{nR{T_1} - nR{T_2}}}{{\gamma - 1}} \cr & = \frac{{nR\left( {{T_1} - {T_2}} \right)}}{{\gamma - 1}} \cr & n = 1,{T_1} = T \Rightarrow \frac{{R\left( {T - {T_2}} \right)}}{{\frac{5}{3} - 1}} = 6R \cr & \Rightarrow {T_2} = \left( {T - 4} \right)K \cr} $$

$$U = a + bPV\,......\left( 1 \right)$$
In adiabatic change,
$$\eqalign{ & dU = - dW = \frac{{nR}}{{\gamma - 1}}\left( {{T_2} - {T_1}} \right) = \frac{{nR}}{{\gamma - 1}}\left( {dT} \right) \cr & \Rightarrow U = \int d U = \frac{{nR}}{{\gamma - 1}}\int d T \cr & {\text{or}}\,\,U = \left( {\frac{{nR}}{{\gamma - 1}}} \right)T + a = \frac{{PV}}{{\gamma - 1}} + a\,......\left( 2 \right) \cr} $$
where $$a$$ is the constant of integration.
Comparing (1) and (2), we get
$$b = \frac{1}{{\gamma - 1}} \Rightarrow \gamma = \frac{{b + 1}}{b}.$$

Molar specific heat of a substance is defined as the amount of heat required to raise the temperature of one gram mole of the substance through a unit degree.
$$\eqalign{ & {\text{As}}\,{\left( {dQ} \right)_p} = \mu {C_p}dT\,\,......\left( {\text{i}} \right)\,\left( {{\text{At}}\,{\text{constant pressure}}} \right) \cr & {\text{and}}\,{\left( {dQ} \right)_V} = \mu {C_V}dT\,\,......\left( {{\text{ii}}} \right)\,\left( {{\text{At}}\,{\text{constant volume}}} \right) \cr & {\text{Given,}}\,{\left( {dQ} \right)_p} = 207\,J \cr & R = 8.3\,J/mol - K \cr & dT = 10\,K \cr & {\text{Putting value in Eq}}{\text{. }}\left( {\text{i}} \right) \cr & 207 = 1 \times {C_p} \times 10 \cr & \therefore {C_p} = 20.7\,J/kg \cr & {\text{As}}\,{C_p} - {C_V} = R = 8.3 \cr & {C_V} = 20.7 - 8.3 = 12.4\,J \cr & \therefore {\left( {dQ} \right)_V} = 1 \times 12.4 \times 10 \cr & = 124\,J \cr} $$

For an adiabatic change $$P{V^\gamma } = {\text{constant}}$$
$${P_1}{V_1}^\gamma = {P_2}{V_2}^\gamma $$
As molar specific heat of gas at constant volume
$$\eqalign{ & {C_V} = \frac{3}{2}R \cr & {C_P} = {C_V} + R = \frac{3}{2}R + R = \frac{5}{2}R; \cr & \gamma = \frac{{{C_P}}}{{{C_V}}} = \frac{{\left( {\frac{5}{2}} \right)R}}{{\left( {\frac{3}{2}} \right)R}} = \frac{5}{3} \cr} $$
∴ From eqn. (1)
$$\eqalign{ & {P_2} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^\gamma }{P_1} = {\left( {\frac{6}{2}} \right)^{\frac{5}{3}}} \times {10^5}\,N/{m^2} \cr & = {\left( 3 \right)^{\frac{5}{3}}} \times {10^5} = 6.19 \times {10^5}\,N/{m^2} \cr} $$
Work done
$$\eqalign{ & = \frac{1}{{1 - \left( {\frac{5}{3}} \right)}}\,\,\left[ {6.19 \times {{10}^5} \times 2 \times {{10}^{ - 3}} - {{10}^{ - 5}} \times 6 \times {{10}^{ - 3}}} \right] \cr & = - \left[ {\frac{{2 \times {{10}^2} \times 3}}{2}\left( {6.19 - 3} \right)} \right] \cr & = - 3 \times {10^2} \times 3.19 \cr & = - 957\,joules\,\,\left[ { - ve\,{\text{sign}}\,{\text{shows}}\,{\text{external}}\,{\text{work}}\,{\text{done}}\,{\text{on}}\,{\text{the}}\,{\text{gas}}} \right] \cr} $$