Heat and Thermodynamics

$\eta = 1 - {{{T_2}} \over {{T_1}}} = 1 - {{273} \over {373}}$

$ = 0.26809$

$ \approx 26.81\% $

Let there be n moles of gas

E_loss = E_gain

${1 \over 2}(nM){v^2} = n{C_v}\Delta T$

${1 \over 2}M{v^2} = {C_v}\Delta T$

here, $\gamma = 1.4 = {7 \over 5}$ i.e. diatomic gas

$\therefore$ ${C_v} = {{5R} \over 2}$

Now, ${1 \over 2}M{v^2} = {{5R} \over 2}\Delta T$

$\Delta T = {{M{v^2}} \over {5R}}$

$6 \times {l^2} = 24$

$ \Rightarrow l = 2$ m

$\therefore$ ${{\Delta V} \over V} = 3 \times {{\Delta l} \over l}$

$ \Rightarrow \Delta V = 3 \times (\alpha \Delta T) \times V$

$ = 3 \times 5 \times {10^{ - 4}} \times 10 \times (8)$

$ = 120 \times {10^{ - 3}}$ m³

$ = 120 \times {10^{ - 3}} \times {10^6}$ cm³

$ = 1.2 \times {10^5}$ cm³

$mL = \Delta Q = ms\Delta T$

$ \Rightarrow m = {{5 \times 0.39 \times {{10}^3} \times 500} \over {335}}$

$ = 2.9$ kg

${{{C_P}} \over {{C_V}}} = \gamma $

${C_V} = \left( {{f \over 2}} \right)R$ and ${C_P} - {C_V} = R$

$ \Rightarrow {{{C_P}} \over C} = {{1 + f/2} \over {f/2}} = 1 + {2 \over f}$

${v_{rms}} = \sqrt {{{3RT} \over M}} $

${v_p} = \sqrt {{{2RT} \over M}} $

$ \Rightarrow {v_{rms}} = \sqrt {{3 \over 2}} {v_p}$

Efficiency $\eta = 1 - {{{T_L}} \over {{T_H}}}$

$ = 1 - {{400} \over {1000}} = 0.6$

$ \Rightarrow 0.6 = {W \over Q}$

$ \Rightarrow W = 0.6Q = 3000$ kcal $=12.6 \times {10^6}$ J

${1 \over 2} \times 1.5 \times {60^2} \times {1 \over 4} = 100 \times 0.42 \times \Delta T$

$\Delta T = {{1.5 \times {{60}^2}} \over {8 \times 100 \times 0.42}} = 16.07^\circ C$

Initially : ${1 \over 4} = 1 - {{300} \over {{T_H}}}$

$ \Rightarrow {T_H} = 400\,K$

Finally : Efficiency becomes ${1 \over 2}$

$ \Rightarrow {1 \over 2} = 1 - {{300} \over {T{'_H}}}$

$ \Rightarrow T{'_H} = 600\,K$

$\Rightarrow$ Temperature of the source increases by 200$^\circ$C.

Thermal current is same so

${{dQ} \over {dt}} = {{\Delta {T_1}} \over {{{{I_1}} \over {{K_1}A}}}} = {{\Delta {T_2}} \over {{{{I_2}} \over {{K_2}A}}}}$

or ${{20} \over {16}} \times K' = {{80} \over 8} \times K$

$ \Rightarrow K' = 8\,K$

For adiabatic process $(\mathrm{A} \rightarrow \mathrm{B})$

$ \begin{aligned} & \mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}^\gamma=\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}^\gamma \\\\ & 10^5 \times(0.8)^{\frac{5}{3}}=3 \times 10^5\left(\mathrm{~V}_{\mathrm{B}}\right)^{\frac{5}{3}} \\\\ & \Rightarrow \mathrm{V}_{\mathrm{B}}=0.8 \times\left(\frac{1}{3}\right)^{0.6}=0.4 \end{aligned} $

Work done in process $\mathrm{A} \rightarrow \mathrm{B}$

$ \begin{aligned} & \mathrm{W}_{\mathrm{AB}}=\frac{\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}-\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}}{\gamma-1} \\\\ & \Rightarrow \mathrm{W}_{\mathrm{AB}}=\frac{10^5 \times 0.8-3 \times 10^5 \times 0.4}{\frac{5}{3}-1} \\\\ & \Rightarrow \mathrm{W}_{\mathrm{AB}}=-60 \mathrm{~kJ}=\Rightarrow\left|\mathrm{W}_{\mathrm{AB}}\right|=60 \mathrm{~kJ} \end{aligned} $

Work done in process $\mathrm{B} \rightarrow \mathrm{C}$ (Isothermal process)

$ \begin{aligned} & \mathrm{W}_{\mathrm{BC}}=\mathrm{nRT} \ell \mathrm{n} \frac{\mathrm{V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{B}}}=\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}} \ell \mathrm{n} \frac{\mathrm{V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{B}}} \\\\ & \Rightarrow \mathrm{W}_{\mathrm{BC}}=3 \times 10^5 \times 0.4 \ell \mathrm{n} \frac{0.8}{0.4} \\\\ & \Rightarrow \mathrm{W}_{\mathrm{BC}}=84 \mathrm{~kJ} \end{aligned} $

Work done in process $\mathrm{C} \rightarrow \mathrm{A}$

$ \mathrm{W}_{\mathrm{CA}}=\mathrm{P} \Delta \mathrm{V}=0 \quad(\because \Delta \mathrm{V}=0) $

So total work done in the process $\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C}$

$ \begin{aligned} & \mathrm{W}_{\mathrm{ABC}}=\mathrm{W}_{\mathrm{AB}}+\mathrm{W}_{\mathrm{BC}}+\mathrm{W}_{\mathrm{CA}}=-60+84+0 \\\\ & \mathrm{~W}_{\mathrm{ABC}}=24 \mathrm{~kJ} \end{aligned} $

So correct options are (B,C,D)

$U = ML - P\Delta V$

$ = {10^{ - 3}} \times 2250 - {10^2}kP \times ({10^{ - 3}} - {10^{ - 6}})$ m³

= 2.25 kJ $-$ 0.1 kJ

= 2.15 kJ

I $\to$ P

(II) ${C_V} = {{5R} \over 2}$ (rigid diatomic)

For isobaric expansion

$V \propto T$

${{{V_1}} \over {{V_2}}} = {{{T_1}} \over {{T_2}}} \Rightarrow {V \over {3V}} = {{500} \over {{T_2}}} \Rightarrow {T_2} = 1500\,K$

$\Delta U = n{C_V}\Delta T = 0.2 \times {{5 \times 8} \over 2} \times (1500 - 500)$ J = 4 kJ

II $\to$ R

(III) Adiabatic expansion $\left( {\gamma = {5 \over 3}} \right)$

${P_1}V_1^\gamma = {P_2}V_2^\gamma \Rightarrow $ (2 kPa) $ \times V_0^{5/3} = {P_2} \times {\left( {{{{V_0}} \over 8}} \right)^{5/3}}$

${P_2} = 64$ kPa

$\Delta U = n{C_V}\Delta T$

$ = {{3nR\Delta T} \over 2} = {3 \over 2}({P_2}{V_2} - {P_1}{V_1}) = {3 \over 2} \times \left( {64 \times {1 \over {3 \times 8}} - 2 \times {1 \over 3}} \right)$

$ = {3 \over 2} \times \left( {{8 \over 3} - {2 \over 3}} \right) = 3$ kJ

III $\to$ T

(IV) For isobaric expaision,

$\Delta U = n{C_V}\Delta T = {7 \over 2}\,nR\Delta T$

$\Delta Q = n{C_P}\Delta T = {9 \over 2}\,nR\Delta T$

${{\Delta U} \over {\Delta Q}} = {7 \over 9}$

$\Delta U = {7 \over 9}\Delta Q = 7$ kJ

IV $\to$ Q

Ring just slip over rod when area of hole is equals to area of cross-section of rod.

$\Rightarrow A_{\text {ring }}$ at $T_2$ temperature $=A_{\text {rod }}$ at $T_2$ temperature

$ \Rightarrow\left[A_0(1+\beta \Delta T)\right]_{\text {ring }}=\left[A_0(1+\beta \Delta T)\right]_{\text {rod }} $

Here,

$ \begin{aligned} & A_0(\text { ring })=9.98 \mathrm{~cm}^2 \\ & A_0(\text { rod })=10 \mathrm{~cm}^2 \\ & \beta_{\text {ring }}=2 \times 17 \times 10^{-6} \mathrm{C}^{-1} \quad(\because \beta=2 \alpha) \\ & \beta_{\text {rod }}=2 \times 11 \times 10^{-6} \mathrm{C}^{-1} \end{aligned} $

So, we have from Eq. (i),

$ \begin{aligned} & 9.98\left(1+2 \times 17 \times 10^{-6} \cdot \Delta T\right) \\ & \quad=10\left(1+2 \times 11 \times 10^{-6} \cdot \Delta T\right) \\ & \text { or } \quad 998+2 \times 998 \times 17 \times 10^{-6} \Delta T \\ & \quad=1000+2 \times 1000 \times 11 \times 10^{-6} \Delta T \\ & \Rightarrow \quad 2 \times 10^{-6} \Delta T(17 \times 998-1000 \times 11) \\ & \quad=1000-998 \\ & \Rightarrow \quad 2 \times 10^{-6} \Delta T(5966)=2 \\ & \Rightarrow \quad \Delta T=\frac{10^6}{5966}=167.61^{\circ} \mathrm{C} \end{aligned} $

Consider statement II, water is formed due to lowering of melting point by increase of pressure (not temperature), when skater skates on ice, increased pressure causes ice to melt and allows skater to move smoothly.

When pressure is released water formed freezes again.

So, statement II is incorrect but statements I and III are correct.

Specific heat of liquids is inversely proportional to the density.

Density of water is maximum at $4^{\circ} \mathrm{C}$.

So, its specific heat is minimum at $4{ }^{\circ} \mathrm{C}$.

Hence, option (c) is incorrect.

In cyclic process, $\Delta U=0$

So, by first law,

$ \begin{aligned} \Delta Q & =\Delta U+\Delta W \\or\,\,\, \Delta Q & =\Delta W \end{aligned} $

As cycle given is clockwise so, work done is positive. Hence, $\Delta Q$ is positive that means heat is absorbed.

Now, $\Delta Q=\Delta W=$ Area of cycle

$ \begin{aligned} & =\frac{1}{2} \times(2 V-V) \times(2 p-p) \\ & =\frac{1}{2} p V \end{aligned} $

Hence, heat released

$ \begin{aligned} & =- \text { Heat absorbed } \\ & =-\frac{1}{2} p V \end{aligned} $

Process is given isothermal, so process equation is $p_1 V_1=p_2 V_2$

$ \begin{array}{lc} \text { Given, } V_2=2 V_1 \\ \Rightarrow p_1 \times V_1=p_2 \times 2 V_1 \\ \Rightarrow p_2=p_1 / 2 \\ \text { but } p_2=p^{\prime} \text { and } p_1=p \\ \therefore p^{\prime}=p / 2 \end{array} $

So, new pressure is $1 / 2$ of initial pressure.

Change in length of combined rod $=$ Change in length of $\operatorname{rod} A+$ Change in length of rod B

$ \begin{aligned}So,\,\, \Delta l & =\Delta l_A+\Delta l_B \\ & =l_A \alpha_A \Delta T+l_B \alpha_B \Delta T \end{aligned} $

$ \begin{aligned}Here,\,\, \Delta T & =T_f-T_i=230-30=200^{\circ} \mathrm{C} \\ l_A & =l_B=50 \mathrm{~cm}=50 \times 10^{-2} \mathrm{~m} \\ \alpha_A & =2 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1} \text { and } \\ \alpha_B & =1 \times 10^{-50}{ }^{\circ} \mathrm{C}^{-1} \end{aligned} $

$ \begin{aligned}So,\,\, \Delta l= & 50 \times 10^{-2} \times 2 \times 10^{-5} \times 200 \\ & +50 \times 10^{-2} \times 1 \times 10^{-5} \times 200 \\ = & 3 \times 10^{-3} \mathrm{~m}=3 \mathrm{~mm} \end{aligned} $

Given, $10 \%$ of total PE of water is used up in heating

$ \begin{array}{cc} \Rightarrow & \text { Heat absorbed }=10 \% \text { of PE } \\ \Rightarrow & m \times s \times \Delta T=\frac{10}{100} \times m g h \\ \Rightarrow & \Delta T=\frac{g \times h}{10 \times s} \\ \Rightarrow & \Delta T=\frac{10 \times 20}{10 \times 4000}=0.5 \times 10^{-2} \\ \Rightarrow & \Delta T=0.005^{\circ} \mathrm{C} \end{array} $

Hence, difference in temperature of water at top and bottom $=0.005^{\circ} \mathrm{C}$

Zeroth law of thermodynamics states about thermal equilibrium. i.e. when two systems are in thermal equilibrium with a third system, then both systems are in equilibrium also with each other.

Relation between volume and temperature in adiabatic expansion is

$ T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1} $

Here, $T_2=\frac{T_1}{2}$ and $V_2=2 V_1$

$ \begin{aligned} \text { So, } & & T_1 V_1^{\gamma-1} & =\frac{T_1}{2}\left(2 V_1\right)^{\gamma-1} \\ \Rightarrow & & 1 & =\frac{1}{2} \cdot 2^{\gamma-1} \\ \Rightarrow & & 2 & =2^{\gamma-1} \\ \Rightarrow & & \gamma-1 & =1 \\ \text { or } & & \gamma & =2 \end{aligned} $

In an isobaric process,

$ \Delta Q=\Delta U+\Delta W \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Here, $\Delta Q=700 \mathrm{~J}$

Gas is given diatomic so, $C_p=\frac{7}{2} R$ and

$ C_V=\frac{5}{2} R $

Heat absorbed by gas is

$ \Delta Q=n C_p \Delta T $

$ \begin{aligned} & 700=n\left(\frac{7}{2} R\right) \cdot \Delta T \\ \Rightarrow & n \Delta T=\frac{700 \times 2}{7 R}=\frac{200}{R}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

So, change in internal energy of gas,

$ \begin{aligned} \Delta U & =n C_V \Delta T=C_V n \Delta T \\ & =\frac{5}{2} R \times \frac{200}{R}=500 \mathrm{~J} \end{aligned} $

[from Eq. (ii)]

So, from Eq. (i), work done is

$ \begin{aligned} \Delta W & =\Delta Q-\Delta U \\ & =700-500 \\ & =200 \mathrm{~J} \end{aligned} $

Given, mass of $\mathrm{CO}_2, m=176 \mathrm{~g}$

Atomic mass of $\mathrm{CO}_2$,

$ M=12+16 \times 2=44 $

∴ Number of mole of $\mathrm{CO}_2$,

$ n=\frac{m}{M}=\frac{176}{44}=4 $

Change in temperature,

$ \Delta T=30-0=30^{\circ} \mathrm{C}=30 \mathrm{~K} $

Heat absorbed, $Q=3600 \mathrm{~J}$

We know that,

$ \begin{aligned} Q & =n s \Delta T \\ \Rightarrow \quad 3600 & =4 \times s \times 30 \\ \Rightarrow \quad s & =\frac{3600}{30 \times 4} \\ & =30 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \end{aligned} $

Latent heat of water, $L_w=80 \mathrm{cal} / \mathrm{g}$ Latent heat of ether, $L_e=90 \mathrm{cal} / \mathrm{g}$ Let $m$ be the mass of ether required, ∴ Heat loss by water $=$ Heat gain by ether

$ \begin{aligned} \Rightarrow & & m_w L_w & =m . L_e \\ \Rightarrow & & 5 \times 80 & =m \times 90 \\ \Rightarrow & & m & =\frac{400}{90}=4.4 \mathrm{~g} \end{aligned} $

which is closest to option (b).

In thermodynamic system, heat and work are not state variables because they depend on the path of connecting states, thus, heat and work are treated as a path functions in thermodynamics. According to first law of thermodynamics,

$ \begin{aligned} & & Q & =\Delta U+W \\ \Rightarrow & & Q-W & =\Delta U \end{aligned} $

Thus, heat $Q$ and work $W$ are modes of energy transfer to a system resulting in change in its internal energy.

In isothermal expansion,

$ \begin{aligned} & p_1 V_1=p_2 V_2 \\ \Rightarrow & p_0 V_1=p_2 .8 V_1 \\ \Rightarrow & p_2=p_0 / 8 \end{aligned} \quad\left[\because V_2=8 V_1\right] $

Now, gas is slowly and adiabatically compressed back to its initial volume. Hence, for this adiabatic compression,

$ \begin{aligned} & p_1=p_0 / 8, V_1=8 \mathrm{~V} \\ & p_2=?, V_2=V \end{aligned} $

∴ Using adiabatic relation,

$ \begin{aligned} & & p_1 V_1^\gamma & =p_2 V_2^\gamma \\ \Rightarrow & & \frac{p_0}{8} \cdot(8 V)^\gamma & =p_2(V)^\gamma \\ \Rightarrow & & \frac{p_0}{8} \cdot 8^\gamma \cdot V^\gamma & =p_2 V^{\prime} \\ \Rightarrow & & \frac{p_0}{8} \cdot 8^\gamma & =p_2 \\ \Rightarrow & & p_2 & =\frac{p_0}{8} \cdot 8^\gamma \\ & & & =\frac{p_0}{8} \cdot(8)^{\frac{4}{3}} \left[\because \gamma=\frac{4}{3}\right]\\ & & & =\frac{p_0}{8} \cdot 16 \\ \Rightarrow & & p_2 & =2 p_0 \end{aligned} $

$ \begin{aligned} &\text { ∴ Average kinetic energy per molecule, }\\ &\begin{array}{ll} & K=\frac{3}{2} k T=\frac{3}{2} \frac{R}{N} T=\frac{3}{2} \frac{p V}{N} \,\,\,\,\,\,\,[\because p V=R T]\\ \Rightarrow & \quad K \propto p \\ \because & \frac{K_2}{K_1}=\frac{p_2}{p_1} \\ \Rightarrow & \frac{K_2}{K_1}=\frac{2 p_0}{p_0} \\ \Rightarrow & \frac{K_2}{K_1}=2 \end{array} \end{aligned} $

Given, atomic mass of Ne,

$ M_1=20.2 \mathrm{u} $

Atomic mass of He ,

$ \begin{aligned} M_2 & =4.0 \mathrm{u} \\ T_2 & =-33^{\circ} \mathrm{C} \\ & =(-33+273) \mathrm{K} \\ & =240 \mathrm{~K} \end{aligned} $

We know, $v_{\text {rms }}=\sqrt{\frac{3 R T}{M}}$

$ \begin{array}{ll} \text { Since, }\left(V_{\text {rms }}\right) \mathrm{He} \\ \Rightarrow \sqrt{\frac{3 R T_2}{M_2}}=\sqrt{\frac{3 R T_1}{M_1}} \\ \Rightarrow \frac{T_2}{M_2}=\frac{T_1}{M_1} \\ \Rightarrow \frac{240}{4}=\frac{T_1}{20.2} \\ \Rightarrow T_1=\frac{20.2 \times 240}{4} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\, =1212 \mathrm{~K} \end{array} $

The volume of metal at $32^{\circ} \mathrm{C}$.

$ \begin{aligned} V_{32 \mathrm{C}} & =\frac{\text { loss of weight }}{\text { specific gravity } \times g} \\ & =\frac{(49-39) g}{1.2 \times g}=8.33 \mathrm{~cm}^3 \end{aligned} $

Similarly, volume of metal at $42^{\circ} \mathrm{C}$.

$ V_{42 \mathrm{C}}=\frac{(49-40) \mathrm{g}}{1 \mathrm{~g}}=9 \mathrm{~cm}^3 $

Now, $V_{42 \mathrm{C}}=V_{32 \mathrm{C}}\left[1+\gamma\left(t_2-t_1\right)\right]$

$ \begin{aligned} \Rightarrow \quad \gamma & =\frac{V_{42{ }^{\circ} \mathrm{C}}-V_{322^{\circ} \mathrm{C}}}{V_{322^{\circ} \mathrm{C}}\left(t_2-t_1\right)}=\frac{9-8.33}{8.33(42-32)} \\ & =\frac{0.66}{83.3}=0.0079 /{ }^{\circ} \mathrm{C} \\ \alpha & =\frac{\gamma}{3}=\frac{0.0079}{3}=\frac{7.9}{3} \times 10^{-3} /{ }^{\circ} \mathrm{C} \\ & \simeq \frac{8}{3} \times 10^{-3} /{ }^{\circ} \mathrm{C} \end{aligned} $

Given, base area of cooking pot,

$ A=0.2 \mathrm{~m}^2 $

Thickness, $l=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}$

Boiling rate, $R=\frac{m}{t}=3 \mathrm{~kg} / \mathrm{min}=\frac{1}{20} \mathrm{~kg} / \mathrm{s}$ Thermal conductivity,

$ K=120 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1} $

Heat of evaporation, $L=2 \times 10^6 \mathrm{~J} / \mathrm{kg}$

The amount of heat flowing into water through the base of cooking pot,

$ Q=\frac{K A\left(T_1-T_2\right) t}{l} $

where, $T_1=$ Temperature of the cooking pot and

$ T_2=\text { Boiling point of water } . $

Heat required to boil water, $Q^{\prime}=m L$

According to given situation, $Q=Q^{\prime}$

$ \begin{aligned} & \frac{K A\left(T_1-T_2\right)}{l} t=m L \\ & \begin{aligned} \Rightarrow T_1-T_2 & =\frac{m L l}{K A t}=\left(\frac{m}{t}\right) \frac{L l}{K A} \\ & =\frac{1}{20} \times \frac{2 \times 10^6 \times 2 \times 10^{-2}}{120 \times 0.2}=83^{\circ} \mathrm{C} \end{aligned} \end{aligned} $

In given process

Pressure $\propto$ Volume $\Rightarrow p=k V$ work done in process,

$ \begin{aligned} \Delta W & =\int_{V_0}^{3 V_0} \rho d V=\int_{V_0}^{3 V_0} k V \cdot d V \\ & =k\left[\frac{V^2}{2}\right]_{V_0}^{3 V_0} \\ & =\frac{k}{2}\left\{9 V_0^2-V_0^2\right\} \\ & =k \cdot 4 V_0^2=\frac{p_0}{V_0} \cdot 4 V_0^2 \end{aligned} $

$or\,\, \Delta W=4 p_0 V_0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..(i)$

As gas is monoatomic, $C_V=\frac{3}{2} R$

Change in internal energy of gas,

$ \Delta U=n C_v\left(T_f-T_i\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..(ii)$

Now, from $p V=n R T$;

$ T_i=\frac{p_0 V_0}{n R} $

and $\quad T_f=\frac{9 p_0 V_0}{n R}$

So, from Eq. (ii)

$ \Delta U=n \cdot \frac{3}{2} R \cdot \frac{8 p_0 V_0}{n R}=12 p_0 V_0 $

By $\Delta Q=\Delta U+\Delta W$, we have Heat absorbed

$ \Delta Q=12 p_0 V_0+4 p_0 V_0=16 p_0 V_0 $

Given, $p_1=p, V_1=V, T_1=T$

$ p_2=p, V_2=4 V, T_2=\frac{T}{2} $

For the adiabatic relation of perfect gas,

$ \begin{aligned} & & T V^{\gamma-1} & =\text { constant } \\ & \Rightarrow & T_1 V_1^{\gamma-1} & =T_2 V_2^{\gamma-1} \\ & \Rightarrow & T V^{\gamma-1} & =\frac{T}{2}(4 V)^{\gamma-1} \end{aligned} $

$ \begin{aligned} \Rightarrow & & V^{\gamma-1} & =\frac{1}{2} \cdot(4)^{\gamma-1} \cdot V^{\gamma-1} \\ \Rightarrow & & 1 & =\frac{1}{2}(4)^{\gamma-1} \Rightarrow 2=(4)^{\gamma-1} \\ \Rightarrow & & 2 & =2^{2(\gamma-1)} \Rightarrow 2(\gamma-1)=1 \\ \Rightarrow & & \gamma-1 & =1 / 2 \\ \Rightarrow & & \gamma & =3 / 2 \end{aligned} $

$ \begin{aligned} &\text { Work done in adiabatic expansion, }\\ &\begin{aligned} W & =\frac{n R}{\gamma-1}\left[T_1-T_2\right] \\ & =\frac{n R}{\frac{3}{2}-1}[T-T / 2] \\ & =2 n R \times \frac{T}{2}=n R T \end{aligned} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad W=p V=\alpha p V \quad & \text { (given) } \\ & {[\because p V=n R T] } \end{aligned} $

$ \therefore \quad \alpha=1 $

$ \text { } \begin{aligned} \left(v_{\mathrm{rms}}\right)_{\mathrm{H}_2} & =\sqrt{\frac{3 R T}{M_{\mathrm{H}_2}}}=\sqrt{\frac{3 R \times 20}{2}} \\ & =\sqrt{30 R} \\ \left(v_{\mathrm{rms}}\right)_{\mathrm{O}_2} & =\sqrt{\frac{3 R T^{\prime}}{M_{\mathrm{O}_2}}}=\sqrt{\frac{3 R T^{\prime}}{32}} \end{aligned} $

According to question,

$ \begin{aligned} & & \left(v_{\mathrm{rms}}\right)_{\mathrm{H}_2} & =\left(v_{\mathrm{rms}}\right)_{\mathrm{O}_2} \\ \Rightarrow & & \sqrt{30 R} & =\sqrt{\frac{3 R T^{\prime}}{32}} \Rightarrow 30 R=\frac{3 R T^{\prime}}{32} \\ \Rightarrow & & 30 \times 32 & =3 T^{\prime} \Rightarrow T^{\prime}=320 \mathrm{~K} \end{aligned} $

$ \begin{aligned} &\alpha=2 \times 10^{-5} \mathrm{~K}^{-1} \Rightarrow T=230^{\circ} \mathrm{C}\\ &\text { We know, } L=L_0[1+\alpha \Delta T]\\ &\begin{aligned} & =5\left[1+2 \times 10^{-5} \times(230-30)\right] \\ L & =5\left[1+\left(2 \times 10^{-5} \times 200\right)\right]=5.02 \mathrm{~cm} \end{aligned} \end{aligned} $