Thermodynamics MCQ Questions & Answers in Heat and Thermodynamics | Physics

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

$${C_V} = \frac{R}{{0.67}} \approx 1.5R = \frac{3}{2}R$$
This is the case of monoatomic gases.
when $${C_V} = \frac{3}{2}R$$

Work done during path $$ACB = {\text{area}}\,ACBB'A'A$$
Work done during path $$BDA = {\text{area}}\,BDAA'B'B$$
∴ Work done during going from $$ACB$$  and then to $$BDA$$  path is $$ = {\text{area}}\,ACBB'A'A - {\text{area}}\,BDAA'B'B$$
$$ = {\text{area}}\,ACBDA$$
NOTE
Net work done in cyclic process is given by area under the cycle.

Relation between Celsius scale and Fahrenheit scale is $$\frac{C}{{100}} = \frac{{F - 32}}{{180}}$$
Putting value of $$F = {140^ \circ }$$
$$\eqalign{ & \therefore \frac{C}{{100}} = \frac{{140 - 32}}{{180}} = 0.6 \cr & \therefore C = {60^ \circ } \cr} $$
Hence, fall in temperature = Temperature of boiling water - final temperature
$${100^ \circ }C - {60^ \circ }C = {40^ \circ }C$$

In adiabatic process, no heat is taken or given by the system i.e.,
$$\Delta Q = 0 \Rightarrow \Delta U = - \Delta W$$
If $$\Delta W$$ is negative (work done on system), then $$\Delta U$$  increases & temperature increases and vice-versa. So work done in adiabatic change in particular gas (ideal gas) depends on change in temperature.

The equation for the line is

$$\eqalign{ & P = \frac{{ - {P_0}}}{{{V_0}}}V + 3\,P \cr & \left[ {{\text{slope}} = \frac{{ - {P_0}}}{{{V_0}}},c = 3\,{P_0}} \right] \cr & P{V_0} + {P_0}V = 3\,{P_0}{V_0}\,\,\,\,\,.....\left( {\text{i}} \right) \cr & {\text{But }}pv = nRT \cr & \therefore \,\,p = \frac{{nRT}}{v}\,\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
From (i) & (ii)
$$\eqalign{ & \frac{{nRT}}{v}{V_0} + {P_0}V = 3\,{P_0}{V_0} \cr & \therefore nRT{V_0} + {P_0}{V^2} = 3\,{P_0}{V_0}\,\,\,\,\,.....\left( {{\text{iii}}} \right) \cr} $$
For temperature to be maximum $$\frac{{dT}}{{dv}} = 0$$
Differentiating e.q. (iii) by $$'v'$$ we get
$$\eqalign{ & nR{V_0}\frac{{dT}}{{dv}} + {P_0}\left( {2\,v} \right) = 3\,{P_0}{V_0} \cr & \therefore \,\,nR{V_0}\frac{{dT}}{{dv}} = 3\,{P_0}{V_0} - 2\,{P_0}V \cr & \frac{{dT}}{{dv}} = \frac{{3\,{P_0}{V_0} - 2\,{P_0}V}}{{nR{V_0}}} = 0 \cr & V = \frac{{3\,{V_0}}}{2} \cr & \therefore \,\,p = \frac{{3\,{P_0}}}{2}\,\,\,\left[ {{\text{From}}\left( {\text{i}} \right)} \right] \cr & \therefore \,\,{T_{\max }} = \frac{{9\,{P_0}{V_0}}}{{4\,nR}}\,\,\left[ {{\text{From}}\left( {{\text{iii}}} \right)} \right] \cr} $$

For adiabatic process, $$dU = - 100\,J$$   which remains same for other processes also.
Let $$C$$ be the heat capacity of 2nd process then
$$\eqalign{ & - \left( C \right)5 = dU + dW = - 100 + 25 = - 75 \cr & \therefore C = 15\,J/K \cr} $$

$$\eqalign{ & W = {W_1} + {W_2} \cr & = 4 \times 1 \times {10^5} + \frac{1}{2} \times 4 \times {10^5} \times 4 \cr & = 12 \times {10^5}\,Joule \cr} $$

$$\eqalign{ & \eta = 1 - \frac{{{T_2}}}{{{T_1}}} \cr & = 1 - \frac{{\left( {273 + 27} \right)}}{{\left( {273 + 627} \right)}} \cr & = 1 - \frac{{300}}{{900}} \cr & = 1 - \frac{1}{3} \cr & = \frac{2}{3} \cr & {\text{But }}\eta = \frac{W}{Q} \cr & \therefore \,\,\frac{W}{Q} = \frac{2}{3} \cr & \Rightarrow \,\,W = \frac{2}{3} \times Q \cr & = \frac{2}{3} \times 3 \times {10^6} \cr & = 2 \times {10^6}\,{{cal}}{\text{.}} \cr & {\text{ = }}2 \times {10^6} \times 4.2\,J \cr & = 8.4 \times {10^6}\,J \cr} $$

$$\eta = 1 - \frac{{{T_2}}}{{{T_1}}}$$
For $$\eta = 1\,\,{\text{or 100% ,}}\,{T_2} = 0\,K.$$
The temperature of 0 $$K$$ (absolute zero) can not be obtained