Thermodynamics MCQ Questions & Answers in Heat and Thermodynamics | Physics

Heat absorbed by gas in three processes is given by
$$\eqalign{ & {Q_{ACB}} = \Delta U + {W_{ACB}} \cr & {Q_{ADB}} = \Delta U \cr & {Q_{AEB}} = \Delta U + {W_{AEB}} \cr} $$
The change in internal energy in all the three cases is same and $${W_{ACB}}$$  is $$ + ve,\,{W_{AEB}}$$   is $$-ve.$$
Hence $${Q_{ACB}} > {Q_{ADB}} > {Q_{AEB}}$$

$$W = \frac{{\pi {r_1}{r_2}}}{2} = \frac{{\pi \times 1 \times 1}}{2} = \frac{\pi }{2}J$$

$$\Delta U = n{C_v}\Delta T = 2 \times {10^3} \times 20 \times 273 = 10.9\,MJ.$$

The work done during the cycle = area enclosed in the curve

Given : Temperature $${T_i} = 17 + 273 = 290\,K$$
Temperature $${T_f} = 27 + 273 = 300\,K$$
Atmospheric pressure, $${P_0} = 1 \times {10^5}\,Pa$$
Volume of room, $${V_0} = 30\,{m^3}$$
Difference in number of molecules, $${N_f} - {N_i} = ?$$
The number of molecules
$$\eqalign{ & \Rightarrow \,\,N = \frac{{PV}}{{RT}}\left( {{N_0}} \right) \cr & \therefore \,\,{N_f} - {N_i} = \frac{{{P_0}{V_0}}}{R}\left( {\frac{1}{{{T_f}}} - \frac{1}{{{T_i}}}} \right){N_0} \cr & = \frac{{1 \times {{10}^5} \times 30}}{{8.314}} \times 6.023 \times {10^{23}}\left( {\frac{1}{{300}} - \frac{1}{{290}}} \right) \cr & = - 2.5 \times {10^{25}} \cr} $$

Here, Coefficient of performance $$\left( \beta \right) = 5$$
$$\eqalign{ & {T_1} = {27^ \circ }C = \left( {27 + 273} \right)K = 300\,K \cr & {\text{As,}}\,\beta = \frac{{{T_2}}}{{{T_1} - {T_2}}} \Rightarrow 5 = \frac{{{T_2}}}{{300 - {T_2}}} \cr & {\text{or}}\,\,1500 - 5{T_2} = {T_2}\,{\text{or}}\,6{T_2} = 1500 \cr & \therefore {T_2} = \frac{{1500}}{6} = 250\,K \cr} $$

For path iaf,
$$\eqalign{ & \Delta U = Q - W \cr & = 50 - 20 \cr & = 30\,\,{{cal}}{\text{.}} \cr} $$
For path ibf,
$$\eqalign{ & W = Q - \Delta U \cr & = 36 - 30 \cr & = 6\,\,{{cal}}{\text{.}} \cr} $$

The adiabatic relation between $$p$$ and $$V$$ for a perfect gas is $$p{V^\gamma } = k\left( {a\,{\text{constant}}} \right)\,......\left( {\text{i}} \right)$$
Again from standard gas equation
$$\eqalign{ & pV = nRT \cr & \Rightarrow V = \frac{{RT}}{p} \cr} $$
Putting in Eq. (i), we get
$$p\frac{{{R^\gamma }{T^\gamma }}}{{{p^\gamma }}} = k$$
or $${p^{1 - \gamma }}{T^\gamma } = \frac{k}{{{R^\gamma }}} = {\text{another}}\,{\text{constant}}$$
i.e. $${p^{1 - \gamma }}{T^\gamma } = {\text{constant}}$$
Comparing two different situations,
$$p_1^1{ - ^\gamma }T_1^\gamma = p_2^1{ - ^\gamma }T_2^\gamma $$
Here, $${p_2} = \left( {\frac{1}{8}} \right){p_1}$$
$$\eqalign{ & {T_1} = {27^ \circ }C = 273 + 27 = 300\,K \cr & {T_2} = ?,\gamma = \frac{5}{3} \cr & \therefore {\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^\gamma } = {\left( {\frac{{{p_1}}}{{{p_2}}}} \right)^{1 - \gamma }} \cr} $$
or $${\left( {\frac{{{T_2}}}{{300}}} \right)^{\frac{5}{3}}} = {\left( 8 \right)^{1 - \frac{5}{3}}} = {\left( 8 \right)^{ - \frac{2}{3}}}$$
$$\eqalign{ & \Rightarrow {T_2} = 130.6\,K \cr & \therefore {T_2} = - {142^ \circ }C \cr} $$

By first law, $$\Delta Q = \Delta U + \Delta W$$
$$\eqalign{ & = \frac{f}{2}nRT + P\int d V = \frac{f}{2}nRT + PV \cr & \Delta Q = \frac{f}{2}nRT + nRT = \frac{3}{2}nRT + nRT = \frac{5}{2}nRT \cr & \therefore {\text{Fraction}} = \frac{{\left( {\frac{3}{2}} \right)nRT}}{{\left( {\frac{5}{2}} \right)nRT}} = \frac{3}{5} \cr} $$

Efficiency of Carnot engine is given by
$$\eta = 1 - \frac{{{T_2}}}{{{T_1}}} = \frac{{{T_1} - {T_2}}}{{{T_1}}}\,.......\left( {\text{i}} \right)$$
Given, $${{T_1}}$$ = temperature of reservoir
$$ = 100 + 273 = 373\,K$$
Given, $${{T_2}}$$ = temperature of sink
$$ = - 23 + 273 = 250\,K$$
Substituting in Eq. (i), we get
$$\eqalign{ & \therefore \eta = \frac{{373 - 250}}{{373}} = \frac{{123}}{{373}} \cr & = \frac{{100 + 23}}{{373}} \cr} $$