Thermodynamics MCQ Questions & Answers in Heat and Thermodynamics | Physics

For path $$iaf,$$
$$Q = 50\,cal$$
$$W = 20\,cal$$
By first law of thermodynamics,
$$\Delta U = Q - W = 50 - 20 = 30\,cal.$$
For path $$ibf$$
$$\eqalign{ & Q' = 36\,cal \cr & W' = ? \cr & {\text{or,}}\,W' = Q' - \Delta U' \cr} $$
Since, the change in internal energy does not depend on the path, therefore
$$\eqalign{ & \Delta U' = 30\,cal \cr & \therefore W' = Q' - \Delta U' = 36 - 30 = 6\,cal. \cr} $$

Change in momentum

$$\eqalign{ & \Delta P = \frac{P}{{\sqrt 2 }}\hat J + \frac{P}{{\sqrt 2 }}\hat J + \frac{P}{{\sqrt 2 }}\hat i - \frac{P}{{\sqrt 2 }}\hat i \cr & \Delta P = \frac{{2P}}{{\sqrt 2 }}\hat J = {I_H}\,\,{\text{molecule}} \cr & \Rightarrow \,\,{I_{{\text{wall}}}} = - \frac{{2P}}{{\sqrt 2 }}\hat J \cr & {\text{Pressure, }}P \cr & = \frac{F}{A} = \frac{{\sqrt 2 P}}{A}n\,\,\left( {\because \,n = {\text{no}}{\text{. of particles}}} \right) \cr & = \frac{{\sqrt 2 \times 3.32 \times {{10}^{ - 27}} \times {{10}^3} \times {{10}^{23}}}}{{2 \times {{10}^{ - 4}}}} \cr & = 2.35 \times {10^3}\,N/{m^2} \cr} $$

$$\eqalign{ & \eta = 1 - \frac{{300}}{{400}} = \frac{{100}}{{400}} = \frac{1}{4} \cr & \eta = \frac{1}{4} \times 100 = 25\% \cr} $$
Hence, it is not possible to have efficiency more than $$25\% .$$

Given, $$\gamma = \frac{{{C_p}}}{{{C_V}}} = 1.5 = \frac{3}{2}$$
$$\therefore {C_V} = \frac{2}{3}{C_p}$$
Again from Mayer's formula
$$\eqalign{ & {C_p} - {C_V} = \frac{R}{J} \cr & \therefore {C_p} - \frac{2}{3}{C_p} = \frac{R}{J} \cr & \Rightarrow \frac{{{C_p}}}{3} = \frac{R}{J} \cr & \Rightarrow {C_p} = \frac{{3R}}{J} \cr} $$

$$\eqalign{ & \Delta {Q_{ACB}} = \Delta {W_{ACB}} + \Delta {U_{ACB}} \cr & \Rightarrow \,\,60\,J = 30\,J + \Delta {U_{ACB}} \cr & \Rightarrow \,\,{U_{ACB}} = 30\,J \cr & \Delta {Q_{ADB}} = \Delta {U_{ADB}} + \Delta {W_{ADB}} \cr & = 10\,J + 30\,J\,\,\left[ {\because \,\,\Delta {U_{ADB}} = \Delta {U_{ACB}} = 30\,J} \right] \cr & = 40\,J \cr} $$

The working of an air conditioner is similar to the working of a refrigerator. An air conditioner removes heat from the room, does some work and rejects the heat to the surroundings. As air conditioner is put in the middle of the room then due to continuous, external work the room will become slightly warmer.

We have given molar specific heat at instant pressure $${C_p} = \frac{7}{2}R$$
Mayer's relation can be written as :
Molar specific heat at constant pressure - Molar specific heat at constant volume = Gas constant,
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}}\,{C_p} - {C_V} = R \Rightarrow {C_V} = {C_p} - R \cr & = \frac{7}{2}R - R = \frac{5}{2}R\left[ {\because {C_p} = \frac{7}{2}R} \right] \cr} $$
Hence, required ratio is $$\gamma = \frac{{{C_p}}}{{{C_V}}} = \frac{{\left( {\frac{7}{2}} \right)R}}{{\left( {\frac{5}{2}} \right)R}} = \frac{7}{5}$$

$${C_p} - {C_v} = {\text{work}}\,{\text{done}}$$

Here $$Q = 0$$  and $$W = 0.$$  Therefore from first law of thermodynamics $$\Delta U = Q + W = 0$$
∴ Internal energy of the system with partition = Internal energy of the system without partition.
$$\eqalign{ & {n_1}{C_v}{T_1} + {n_2}{C_v}{T_2} = \left( {{n_1} + {n_2}} \right){C_v}\,T \cr & \therefore \,\,T = \frac{{{n_1}{T_1} + {n_2}{T_2}}}{{{n_1} + {n_2}}} \cr & {\text{But }}{n_1} = \frac{{{P_1}{V_1}}}{{R{T_1}}}\,{\text{and }}{n_2} = \frac{{{P_2}{V_2}}}{{R{T_2}}} \cr & \therefore \,\,T = \frac{{\frac{{{P_1}{V_1}}}{{R{T_1}}} \times {T_1} + \frac{{{P_2}{V_2}}}{{R{T_2}}} \times {T_2}}}{{\frac{{{P_1}{V_1}}}{{R{T_1}}} + \frac{{{P_2}{V_2}}}{{R{T_2}}}}} \cr & = \frac{{{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)}}{{{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}} \cr} $$

Isochoric proceess $$dV = 0$$
$$W = 0$$  proceess 1
Isobaric : $$W = P\Delta V = nR\Delta T$$
Adiabatic $$\left| W \right| = \frac{{nR\Delta T}}{{\gamma - 1}}$$
$$0 < \gamma - 1 < 1$$
As work done in case of adiabatic process is more so process 3 is adiabatic and process 2 is isobaric.