Thermodynamics MCQ Questions & Answers in Heat and Thermodynamics | Physics

Heat absorbed in a thermodynamic process is given by $$\Delta Q = \Delta U + \Delta W.$$
Here $$\Delta U$$  is same for all the three processes as it depends only on initial and final states.
$$\eqalign{ & {\text{But}}\,\Delta {W_I} = + Ve,\Delta {W_{II}} = 0,\Delta {W_{III}} = - ve \cr & \therefore \Delta {Q_I} > \Delta {Q_{II}} \cr} $$

$$PV$$  = constant. Differentiating,
$$\eqalign{ & \frac{{P\,dV}}{{dP}} = - V;\beta \cr & = - \left( {\frac{1}{V}} \right)\left( {\frac{{dV}}{{dP}}} \right) \cr & = \left( {\frac{1}{P}} \right) \cr & \Rightarrow \,\,\beta \times P = 1 \cr} $$
∴ Graph between $$\beta $$ and $$P$$ will be a rectangular hyperbola.

According to first law of thermodynamics
$$\eqalign{ & Q = \Delta U + W \cr & {\text{Given:}}\,Q = 2\,kcal = 2000 \times 4.2 = 8400\,J \cr & W = 400\,J \cr & \therefore \Delta U = Q - W = 8400 - 400 = 8000\,J \cr} $$

$$\eqalign{ & {\eta _1} = 1 - \frac{{{T_2}}}{{{T_1}}} \Rightarrow \frac{1}{6} = 1 - \frac{{{T_2}}}{{{T_1}}} \Rightarrow \frac{{{T_2}}}{{{T_1}}} = \frac{5}{6}\,......\left( {\text{i}} \right) \cr & {\eta _2} = 1 - \frac{{{T_2} - 62}}{{{T_1}}} \Rightarrow \frac{1}{3} = 1 - \frac{{{T_2} - 62}}{{{T_1}}}\,......\left( {{\text{ii}}} \right) \cr} $$
On solving Eqs. (i) and (ii)
$${T_1} = 372\,K\,{\text{and}}\,{T_2} = 310\,K$$

$$\eqalign{ & W = P\left( {\Delta V} \right) = m\left( {{C_p} - {C_\nu }} \right)\Delta T \cr & = 4\left( {0.219 - 0.157} \right) \times 4200 \times \left( {120 - 20} \right) \cr & \simeq 104 \times {10^3}\,J \cr} $$

The solution of this question can be understood by plotting a $$p-V$$  graph for the compression of a gas isothermally and adiabatically simultaneously to half of its initial volume. i.e.

Since, the isothermal curve is less steeper than the adiabatic curve. So, area under the $$p-V$$  curve for adiabatic process has more magnitude than isothermal curve. Hence, work done in adiabatic process will be more than in isothermal process.

In the first process $$W$$ is $$+ ve$$  as $$\Delta V$$ is positive, in the second process $$W$$ is $$-ve$$  as $$\Delta V$$ is $$-ve$$  and area under the curve of second process is more
$$\therefore {\text{Net}}\,{\text{Work}}\, < 0\,{\text{and}}\,{\text{also}}\,{P_3} > {P_1}.$$

$$\eqalign{ & \eta = 1 - \frac{{{T_2}}}{{{T_l}}}\,{\text{or}}\,\frac{{50}}{{100}} = 1 - \frac{{500}}{{{T_1}}} \Rightarrow {T_1} = 1000\,K \cr & {\text{Also,}}\,\,\frac{{60}}{{100}} = 1 - \frac{{{T_2}}}{{1000}} \Rightarrow {T_2} = 400\,K \cr} $$

If $${n_1}$$ moles of adiabatic exponent $${\gamma _1}$$ is mixed with $${n_2}$$ moles of adiabatic exponent $${\gamma _2}$$ then the adiabatic component of the resulting mixture is given by
$$\eqalign{ & \frac{{{n_1} + {n_2}}}{{\gamma - 1}} = \frac{{{n_1}}}{{{\gamma _1} - 1}} + \frac{{{n_2}}}{{{\gamma _2} - 1}} \cr & \frac{{1 + 1}}{{\gamma - 1}} = \frac{1}{{\frac{7}{5} - 1}} + \frac{1}{{\frac{5}{3} - 1}} \cr & \therefore \,\,\frac{2}{{\gamma - 1}} = \frac{5}{2} + \frac{3}{2} \cr & = 4 \cr & \therefore 2 = 4\gamma - 4 \cr & \Rightarrow \,\,\gamma = \frac{6}{4} \cr & = \frac{3}{2} \cr} $$

Change in internal energy of a gas having atomicity $$\gamma $$ is given by
$$\Delta U = \frac{1}{{\left( {\gamma - 1} \right)}}\left( {{p_2}{V_2} - {p_1}{V_1}} \right)$$
Given, $${V_1} = V,{V_2} = 2V$$
So, $$\Delta U = \frac{1}{{\gamma - 1}}\left[ {p \times 2V - p \times V} \right]$$
$$\eqalign{ & = \frac{1}{{\gamma - 1}} \times pV \cr & = \frac{{pV}}{{\gamma - 1}} \cr} $$
NOTE
The internal energy of an ideal gas depends only on its absolute temperature $$\left( T \right)$$  and is directly proportional to $$T.$$