Thermodynamics MCQ Questions & Answers in Heat and Thermodynamics | Physics

From first law of thermodynamics $$\Delta Q = \Delta U + \Delta W$$
For adiabatic process, $$\Delta Q = 0$$
$$\eqalign{ & \because \Delta Q = 0 \cr & {\text{So,}}\,\,\Delta U = - \Delta W \cr} $$

1^st process is isothermal expansion which is only correct shown in option (D)
2^nd process is isobaric compression which is correctly shown in option (D)

$$\eqalign{ & Q = {Q_1} + {Q_2} + {Q_3} + {Q_4} \cr & = 6000 - 5500 - 3000 + 3500 = + 1000\,J \cr & W = {W_1} + {W_2} + {W_3} + {W_4} \cr & = 2500 - 1000 - 1200 + x = + 300 + x \cr} $$
In cyclic process, $$\Delta U = 0$$
Now, $$Q = \Delta U + W$$
$$\eqalign{ & {\text{or}}\,\,1000 = 0 + \left( {300 + x} \right) \cr & \therefore x = 700\,J \cr & \eta = \frac{W}{{{Q_1} + {Q_4}}} \cr & = \frac{{1000}}{{6000 + 3500}} = 10.5\% \cr} $$

$$\eqalign{ & {W_{ABC}} = \frac{{\pi {r^2}}}{2} = \frac{{\pi {{\left( 6 \right)}^2}}}{2} = 18\pi \cr & {W_{DEF}} = \frac{{\pi \times {3^2}}}{2} + \frac{{\pi \times {3^2}}}{4} + \left( {15 - 12} \right) \times 18 \cr & = 6.75\pi + 54 \cr & {W_{DEF}} > {W_{ABC}} \cr} $$

Change in internal energy for cyclic process $$\left( {\Delta U} \right) = 0.$$
For process $$a \to b,\,\left( {P{\text{ - constant}}} \right)$$
$$\eqalign{ & {W_{a \to b}} = P\Delta V \cr & = nR\Delta T = - 400\,R \cr} $$
For process $$b \to c,\,\,\left( {T{\text{ - constant}}} \right)$$
$${W_{b \to c}} = - 2R\left( {300} \right)\ln 2$$
For process $$c \to d,\,\left( {P{\text{ - constant}}} \right)$$
$${W_{c \to d}} = + 400\,R$$
For process $$d \to a,\,\left( {T{\text{ - constant}}} \right)$$
$${W_{d \to a}} = + 2R\left( {500} \right)\ln 2$$
Net work $$\left( {\Delta W} \right) = {W_{a \to b}} + {W_{b \to c}} + {W_{c \to d}} + {W_{d \to a}}$$
$$\eqalign{ & \Delta W = 400\,R\ln 2 \cr & \therefore dQ = dU + dW,\,{\text{first law of thermodynamics}} \cr & \therefore dQ = 400\,R\ln 2. \cr} $$

If $${T_1}$$ is temperature of source and $${T_2}$$ the temperature of sink, the efficiency of engine
$$\eqalign{ & \eta = \frac{{{\text{Work done}}\left( W \right)}}{{{\text{Heat taken}}\left( {{Q_1}} \right)}} = 1 - \frac{{{T_2}}}{{{T_1}}} \cr & \therefore 1 - \frac{{{T_2}}}{{{T_1}}} = \frac{1}{6}\,......\left( {\text{i}} \right) \cr} $$
When temperature of sink is reduced by $${62^ \circ }C,$$  then temperature of sink
$$\eqalign{ & {{T'}_2} = {T_2} - 62 \cr & \therefore \eta ' = 1 - \frac{{{{T'}_2}}}{{{T_1}}} \cr} $$
As according to question efficiency becomes double
So, $$\eta ' = 2\eta = \frac{2}{6} = \frac{1}{3}$$
$$\therefore \frac{1}{3} = 1 - \frac{{{T_2} - 62}}{{{T_1}}}\,......\left( {{\text{ii}}} \right)$$
From Eq. (i) $$\frac{{{T_2}}}{{{T_1}}} = \frac{5}{6}\,......\left( {{\text{iii}}} \right)$$
From Eq. (ii) $$\frac{{{T_2} - 62}}{{{T_1}}} = \frac{2}{3}\,......\left( {{\text{iv}}} \right)$$
Dividing Eq. (iii) by Eq. (iv)
$$\eqalign{ & \frac{{{T_2}}}{{{T_2} - 62}} = \frac{5}{4} \cr & \Rightarrow 4{T_2} = 5{T_2} - 310 \cr & \Rightarrow {T_2} = 310\,K \cr} $$
and from Eq. (iii), we have
$$\frac{{310}}{{{T_1}}} = \frac{5}{6} \Rightarrow {T_1} = 372\,K$$
Hence, $${T_1} = 372\,K = 372 - 273 = {99^ \circ }C$$
and $${T_2} = 310\,K = 310 - 273 = {37^ \circ }C$$
As kinetic energy of a gas depends on its atomicity.

$$\eqalign{ & \tau = \frac{1}{{\sqrt 2 \pi {d^2}\left( {\frac{N}{V}} \right)\sqrt {\frac{{3RT}}{M}} }} \cr & t\mu \frac{V}{{\sqrt T }} \cr & {\text{As, }}T{V^{\gamma - 1}} = K \cr & {\text{So, }}\tau \, \propto \,{V^{\gamma + \frac{1}{2}}} \cr & {\text{Therefore, }}q = \frac{{g + 1}}{2} \cr} $$

Thermodynamics concerned with $$\Delta H,\Delta U$$   and $$\Delta W.$$

As we know that for polytropic process of index $$\alpha $$ specific heat capacity $$ = {C_V} + \frac{R}{{1 - \alpha }}$$
$$\eqalign{ & \because {\text{Process,}}\,\,p{V^3} = {\text{constant}} \Rightarrow \alpha = 3 \cr & \therefore C = {C_V} + \frac{R}{{1 - \alpha }} = \frac{{fR}}{2} + \frac{R}{{1 - 3}} \cr & {\text{where,}}\,\,{C_V} = \frac{{fR}}{2} = \frac{{3R}}{2} \cr} $$
For monatomic gas, $$f = 3 = \frac{{3R}}{2}$$
$$ \Rightarrow C = \frac{{3R}}{2} - \frac{R}{2} = R$$

As we know, $${C_P} - {C_V} = R$$   where $${C_P}$$ and $${C_V}$$ are molar specific heat capacities
or, $${C_P} - {C_V} = \frac{R}{M}$$
For hydrogen $$\left( {M = 2} \right){C_P} - {C_V} = a = \frac{R}{2}$$
For nitrogen $$\left( {M = 28} \right){C_P} - {C_V} = b = \frac{R}{28}$$
$$\therefore \,\,\frac{a}{b} = 14\,\,{\text{or, }}a = 14\,b$$