Thermodynamics MCQ Questions & Answers in Heat and Thermodynamics | Physics

The specific heat of gas at constant volume in terms of degree of freedom $$n$$ is $${C_V} = \frac{n}{2}R$$
Also $${C_p} - {C_V} = R$$
So $${C_p} = \frac{n}{2}R + R = R\left( {1 + \frac{n}{2}} \right)$$
Now, $$\gamma = \frac{{{C_p}}}{{{C_V}}} = \frac{{R\left( {1 + \frac{n}{2}} \right)}}{{\frac{n}{2}R}} = \frac{2}{n} + 1$$

In an isochoric process, no work is done on or by the gas. $$V$$ is constant

Change in internal energy do not depend upon the path followed by the process. It only depends on initial and final states i.e.,
$$\Delta {U_1} = \Delta {U_2}$$

Given that $$dQ = - 30\,J$$   and $$dW = - 10\,J,$$   $${E_i} = 40\,J$$   and let final internal energy $$ = {E_f}$$
So, $$dQ = {E_f} - {E_i} + dW$$
$$\eqalign{ & - 30 = {E_f} - 40 - 10 \cr & {E_f} = 20\,J \cr} $$

The $$p-V$$  diagram for isobaric, isothermal and adiabatic processes of an ideal gas is shown in graph below

In thermodynamics, for some change in volume, the work done is maximum for the curve having maximum area enclosed with the volume axis. Area enclosed by the curve $$ \propto \left( {{\text{Slope of curve}}} \right)$$
NOTE
$$\eqalign{ & {\left( {{\text{Slope}}} \right)_{{\text{isobaric}}}} < {\left( {{\text{Slope}}} \right)_{{\text{isothermal}}}} < {\left( {{\text{Slope}}} \right)_{{\text{adiabatic}}}} \cr & {\left( {{\text{Area}}} \right)_{{\text{isobaric}}}} > {\left( {{\text{Area}}} \right)_{{\text{isothermal}}}} > {\left( {{\text{Area}}} \right)_{{\text{adiabatic}}}} \cr} $$
Hence, work done is maximum in isobaric process.
$$\eqalign{ & {\left( {{\text{Slope}}} \right)_{{\text{adiabatic}}}} = - \gamma \left( {\frac{p}{V}} \right) \cr & {\text{and}}\,\,{\left( {{\text{Slope}}} \right)_{{\text{isothermal}}}} = - \frac{p}{V} \cr} $$
$$\therefore {\left( {{\text{Slope}}} \right)_{{\text{adiabatic}}}} = \gamma \times {\left( {{\text{Slope}}} \right)_{{\text{isothermal}}}}$$
Slope of adiabatic curve is always steeper than that of isothermal curve.

$$\eqalign{ & VT = {\text{constant}} \cr & \therefore PV = nRT \cr & V = \left[ {\frac{{nRT}}{P}} \right] \cr & \therefore \frac{{nRT}}{P}.T = {\text{constant}} \cr & \Rightarrow P \propto {T^2}. \cr} $$

$$\eqalign{ & Q = mC\Delta T = 1.5 \times 0.12 \times 4200 \times \left( {400 - 25} \right) \cr & = 2.83 \times {10^5}\,J \cr & W = P\left( {\Delta V} \right) = P\left( {V\gamma \Delta T} \right) \cr & = {10^5} \times {\left( {5 \times {{10}^{ - 2}}} \right)^3} \times 3.5 \times {10^{ - 5}} \times 375 = 0.164\,J \cr & {\text{Thus}}\,\,Q = \Delta U + W \cr & {\text{or}}\,\,2.83 \times {10^5} = \Delta U + 0.164;\Delta U = 282\,kJ \cr} $$

$$\eqalign{ & P{V^{\frac{3}{2}}} = {\text{constant}} \cr & PV = nRT \cr & \Rightarrow {V^{\frac{1}{2}}}T = {\text{const}}{\text{.}} \cr & \Rightarrow {\left( {\frac{V}{2}} \right)^{\frac{1}{2}}}T' = {\text{const}}{\text{.}} \cr & \Rightarrow T' = T\sqrt 2 \cr} $$

Internal energy of an ideal gas depends only on the temperature.

Internal energy depends only on initial and final state
So, $$\Delta {U_A} = \Delta {U_B}$$
$$\eqalign{ & {\text{Also }}\Delta Q = \Delta U + \Delta W \cr & \therefore \,\,{W_A} > {W_B} \cr & \Rightarrow \,\,\Delta {Q_A} > \Delta {Q_B} \cr} $$
[Area under $$P - V$$ ; graph gives the work done. ]