Physics M.C.Q (1 Marks)

$\therefore \quad$ Work done by gas along path $b c$ is zero.

$\% \eta=\left(1-\frac{T_{\text {sink }}}{T_{\text {source }}}\right) \times 100$

$T_{\text {source }}=327^{\circ}\,C =600\,K$

$50=\left(1-\frac{T_{\text {sink }}}{600}\right) \times 100$

$\frac{1}{2}=1-\frac{T_{\text {sink }}}{600}$

$T _{\text {Sink }}=300\,K$

So temp. of sink is ${ }^{\circ} C =300-2763=27^{\circ}\,C$

$2$: Adiabatic

$3$ : Isothermal

$4$: Isobaric

$\Rightarrow \frac{ nRT }{ V } V ^2= C$

$\Rightarrow TV =\text { constant }$

$\therefore V _2 > V _1 \Rightarrow T _1 > T _2$

$\eta=\left(1-\frac{T_{2}}{T_{1}}\right) \times 100 \%$

So efficiency depends on temperature of source $\left(T_{1}\right)$ and temperature of $\sin k\left(T_{2}\right)$

$\Delta Q=0$

$\Delta Q = \Delta U + \Delta W$

$ \Rightarrow 54 \times 4.18 = \Delta U + 1.013 \times {10^5}\left( {167.1 \times {{10}^{ - 6}} - 0} \right)$

$ \Rightarrow \Delta U = 208.7\,J$

$\therefore \,dQ = n{C_p}dT;\,where\,{C_p}\,is\,specific\,heat\,at\,constant\,pressure.$

or        $dQ = n\left( {\frac{5}{2}R} \right)dT$

$Also,\,dW = PdV = nRdT\,\left( {PV = nRT} \right)$

Required ratio$ = \frac{{dW}}{{dQ}} = \frac{{nRdT}}{{n\left( {\frac{5}{2}R} \right)dT}} = \frac{2}{5}$

$\eta  = \left( {1 - \frac{{{T_2}}}{{{T_1}}}} \right)$

Freezing point of water $ = {0^ \circ }C = 273\,K$

Boiling point of water$ = {100^ \circ }C = \left( {100 + 273} \right)K$

$ = 373\,K$

$T_2$ Sink temperature$=273 K$

$T_1$ Source temperature $=373 K$

$\% \eta  = \left( {1 - \frac{{{T_2}}}{{{T_1}}}} \right) \times 100 = \left( {1 - \frac{{273}}{{373}}} \right) \times 100$

$ = \left( {\frac{{100}}{{373}}} \right) \times 100 = 26.8\% $

$VT=$constant

$V(p V)=$ constant

$\therefore p V^{2}=$ constant

In the process $p V^{x}=$ constant, molar heat capacity is

$C=\frac{R}{\gamma-1}+\frac{R}{1-x}$

Here, $x=2$

$\therefore C=\frac{R}{\gamma-1}+\frac{R}{1-2}=\left(\frac{2-\gamma}{\gamma-1}\right) R$

Now, $Q=n C \Delta T$

$=(1)\left(\frac{2-\gamma}{\gamma-1}\right) R \Delta T$

$=\left(\frac{2-\gamma}{\gamma-1}\right) R \Delta T$

$\therefore \,\,\,\,process\,I \to Isochoric;\,P \to C$

As slope of curve $II$ is more than the slope of curve $III$.

$process\,II \to Adiabatic\,and\,processIII \to Isothermal$

$\therefore \,\,\,Q \to A,R \to D$

In process $IV$, pressure is constant

Process $IV \to Isobaric;\,S \to B$

$W=\frac{1}{2} \text { base } \times \text { height }+ \text { Area of rectangular }$

$=\frac{1}{2}\left(V_2-V_1\right) \times\left(P_1-P_2\right)+\left(V_2-V_1\right) P_2$

$=\left(V_2-V_1\right)\left[\frac{P_1}{2}-\frac{P_2}{2}+P_2\right]$

$=\left(V_2-V_1\right)\left[\frac{P_1}{2}+\frac{P_2}{2}\right]$

$=\frac{1}{2}\left(P_1+P_2\right)\left(V_2-V_1\right)$

Room temperature $ = {t_1}{\,^ \circ }C$

For refrigerator,

$\frac{{Heat\,given\,to\,high\,temperature\,\left( {{Q_1}} \right)}}{{Heat\,taken\,from\,lower\,temperature\,\left( {{Q_2}} \right)}} = \frac{{{T_1}}}{{{T_2}}}$

$\frac{{{Q_1}}}{{{Q_2}}} = \frac{{{t_1} + 273}}{{{t_2} + 273}}$

$ \Rightarrow \frac{{{Q_1}}}{{{Q_1} - W}} = \frac{{{t_1} + 273}}{{{t_2} + 273}}\,\,or\,\,1 - \frac{W}{{{Q_1}}} = \frac{{{t_2} + 273}}{{{t_1} + 273}}$

$or\,\,\frac{W}{Q_1} = \frac{{{t_1} - {t_2}}}{{{t_1} + 273}}$

The amount of heat delivered to the room for each joule pf electrical energy $\left( {W = 1\,J} \right)$

${Q_1} = \frac{{{t_1} + 273}}{{{t_1} - {t_2}}}$

${Q_2} = 600\,cal\,per\,second$

Coefficient of performance, $\alpha  = \frac{{{T_2}}}{{{T_1} - {T_2}}}$

$ = \frac{{277}}{{303 - 277}} = \frac{{277}}{{26}}$

Also,$\alpha  = \frac{{{Q_2}}}{W}$

$\therefore $ $Work\,to\,be\,done\,per\,second=power\,required$

$ = W = \frac{{{Q_2}}}{\alpha } = \frac{{26}}{{277}} \times 600\,cal\,per\,second$

$ = \frac{{26}}{{277}} \times 600 \times 4.2\,J\,per\,second = 236.5\,W$

$On\,P - V\,diagram,$

$Area\,under\,adiabatic\,curve>Area\,under\,isothermal\,curve,$

So compressing the gas through adiabatic process will require more work to be done.

$P \propto T$
i.e., pressure will be doubled if temperature is doubled
 $P = 2{P_0}$
Now let F be the tension in the wire. Then equilibrium of any one piston gives
$F = (P - {P_0})A = (2{P_0} - {P_0})A = {P_0}A$

$P V=n R T$

$P \propto n T(\text { at constant } V)$

thus, $P-T$ curves are linear but with different slopes depending on $n .$

$\Delta Q=n C_{p} \Delta T$

$\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$

From first law of thermodynamics

$\Delta Q=\Delta U+\Delta W$

$\Delta \mathrm{W}=\Delta Q+\Delta \mathrm{U}$

$\Delta \mathrm{W}=\mathrm{nR} \Delta \mathrm{T}$

As $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$

$\frac{\Delta W}{\Delta Q}=\frac{n R \Delta T}{n C_{P} \Delta T}=\frac{R}{C_{P}}=\frac{R}{\frac{5}{2} R}=\frac{2}{5}$

$V^{1.33} =$ constan

$V =$ constant

at constant pressure $Q=n C_{p} \Delta T$

$\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$

$\Rightarrow \frac{W}{Q}=\frac{n \Delta T\left(C_{p}-C_{V}\right)}{n C_{p} \Delta T}=\frac{C_{p}-C_{V}}{C_{p}}=\frac{\gamma}{\gamma-1}$

adiabatic process, $\mathrm{Q}=0$ For isothermal process

${Q_{rejected}}\, =  - W = nR{T_0}\ell n\left( {{V_0}/\frac{{{V_0}}}{2}} \right) = nR{T_0}\ell n\,2$

$=0.693 \mathrm{nRT}_{0}$

For isobaric process

$Q_{\text {rejected }}=-n C_{p} \Delta T=-n\left(\frac{f}{2}+1\right) n R\left(-\frac{T_{0}}{2}\right)$

$=\left(0.5+\frac{f}{4}\right) n R T_{0}>0.693 n R T_{0}$

It is clear that more heat is rejected in isobaric process.

$=10^{5}(1671-1) \times 10^{-6} \mathrm{J}$

$=\frac{10^{5} \times 1670 \times 10^{-6}}{4.2} \mathrm{cal}$

$W=\frac{1670 \times 10^{-1}}{4 \cdot 2}$

$=40 \mathrm{cal}$

Cylinder $A$

Free piston i.e., at constant pressure

Cylinder $B$

Fixed piston i.e., at constant volume

$\Delta Q=\Delta U$

$n C_P \Delta T=n C_V \Delta T$

$C_P T_0=C_V(\Delta T)^{\prime}$

$\Delta T^{\prime}=\frac{C_P}{C_V} T_0=\gamma T_0=\frac{5}{3} T_0$

In the process $A \rightarrow B$

Pressure is constant.

$P V=n R T$

So $V \propto T$

and volume is increasing so temperature also increases.

$dW = - \frac{{PV}}{2}$

$ = \left( {4 \times \frac{3}{2}RT + 2 \times \frac{5}{2}RT} \right) - \left( {4 \times \frac{5}{2}RT} \right) = RT$
$Note : \,(a)\, 2$ moles of diatomic gas becomes $4$ moles of a monoatomic gas when gas dissociated into atoms.
$(b)$ Internal energy of $\mu $ moles of an ideal gas of degrees of freedom $F$ is given by $U = \frac{f}{2}\mu RT$
$F = 3$ for a monoatomic gas and $5$ for diatomic gas.

(i) Isothermal expansion $(PV = {\rm{constant)}}$ at temperature ${T_A}$ to twice the initial volume ${V_A}$

Compression at constant pressure $\frac{{{P_A}}}{2}$ to original volume ${V_A}\,(i.e.\,V \propto T)$

(iii)Isochoric process (at volume ${V_A}$) to initial condition $(i.e.\,P \propto T)$

$\mathrm{BC}$ - isochoric process

$\mathrm{CA}-$ isothermal process

$\left(\rho_{1} {g} 5\right)(2 \mathrm{V})=\left[\left(\rho_{1} g 75\right)+\rho_{2}{g} h\right] V$

$\Rightarrow \rho_{1} \mathrm{g} 75=\rho_{2} {g} \mathrm{h}$

$\Rightarrow \mathrm{h}=75 \times \frac{\rho_{1}}{\rho_{2}}=\frac{75 \times 40}{3}=1000 \mathrm{cm}=10 \mathrm{m}$

For a straight $P-V$ graph line $P \propto V$

If pressure increases, volume increases then $T$ also increases $[P V \propto T]$

So $\Delta T \neq 0$

Volume increasing so work is positive, $W > 0$

and temperature also increasing so $\Delta Q > 0$

$\because \Delta Q=\Delta U+\Delta W$

So $\Delta U > 0$

From $A \longrightarrow B$, volume increasing, pressure constant

$B \longrightarrow C$, Pressure $\propto \frac{1}{\text { Volume }} \Rightarrow$ Temperature constant

Same for $D \longrightarrow A$

$C \longrightarrow D$ pressure decreasing, volume constant

So $P \propto T$

$T_3 > T_2 > T_1$

So from $C \rightarrow D$

Temperature first increases then decreases.

As the bubble rises the pressure gets reduced for constant temperature, if $P$ is the standard atmospheric pressure, then $(P+\rho g h) V_0=P V$ or $V = V _0\left(1+\frac{\rho gh }{ P }\right)$

==>$\frac{{{T_2}}}{{300}} = {\left( {\frac{4}{1}} \right)^{\frac{{(1 - 1.4)}}{{1.4}}}}$

==>${T_2} = 300{(4)^{ - \frac{{0.4}}{{1.4}}}}$

Clearly, ${\left( {\frac{{dP}}{{dV}}} \right)_{{\rm{adiabatic}}}} = \gamma {\left( {\frac{{dP}}{{dV}}} \right)_{{\rm{Isothermal }}}}$

==> $\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^\gamma }$

==> ${T_2} = {T_1}{\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^\gamma }$

$ \Rightarrow {T_2} = 300{\left( {\frac{1}{2}} \right)^{0.4}} = 227.36\;K$

$ \Rightarrow \frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{1}{8}} \right)^{\frac{{1.5 - 1}}{{1.5}}}} = {\left( {\frac{1}{8}} \right)^{\frac{1}{3}}} = \frac{1}{2}$
$ \Rightarrow {T_2} = \frac{{{T_1}}}{2} = \frac{{300}}{2} = 150K$.

==> ${T_2} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}} \times {T_1}$
==> ${T_2} = {\left( {\frac{V}{{V/4}}} \right)^{1.4 - 1}} \times 300$ $ = 300 \times {(4)^{0.4}}K$

==>${E_\theta } = \frac{{{E_\varphi }}}{\gamma }$
==> ${E_\theta } = \frac{{2.1 \times {{10}^5}}}{{1.4}}$$ = 1.5 \times {10^5}N/{m^2}$

$P{V^{3/2}}$= constant

or $PV = {P_2} \times 4V$

$\therefore {P_2} = \frac{P}{4}$

In adiabatic process

${P_2}{V_2}^\gamma = {P_3}{V_3}^\gamma $==> $\frac{P}{4} \times {(4V)^{1.5}} = {P_2}{V^{1.5}}$==> ${P_3} = 2P$

$\frac{{dP}}{{dV}} = - \gamma \frac{P}{V}$==> |Slope| $\propto \gamma$

From the given graph $(Slope)_2 > (Slope)_1$ ==> ${\gamma _2} > {\gamma _1}$ 

therefore $1$ should correspond to $O_2 (\gamma$ = 1.4) and $2$ should correspond to $He$ ($\gamma = 1.66$)

==> $\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}}$==> ${T_2} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}} \times {T_1}$

==> ${T_2} = {\left( {\frac{1}{{81}}} \right)^{1.25 - 1}} \times 273$$ = {\left( {\frac{1}{{81}}} \right)^{0.25}} \times 273$

$ = \frac{{273}}{3} = 91K = \,-182°C$

==> ${V_2} < \frac{{{V_1}}}{2}$

==> ${T_2} = \frac{{{T_1}}}{{1.25}} = \frac{{(273 + 27)}}{{1.25}} = 238\,K = - \,34.8^\circ C$

${P_C}(V) = P(2V)$

==> ${P_A}\,:\,{P_B}\,:\,{P_C} = {(2)^{3/2}}:\,1\,\,:\,\,2 = 2\sqrt 2 \,:\,1\,:\,\,2$

$P V^{\gamma}=$ constant

or

$\frac{P_{1}}{P_{2}}=\left(\frac{V_{2}}{V_{1}}\right)^{\gamma}$

or

$P_{2}=2^{\gamma} P_{1}$

Now, for a monoatomic gas, the value of $\gamma$ is the highest

Thus, for the same change in volume, the monoatomic gas will have the maximum pressure.

in $B$ temperature remains constant

in $C \& D$ temperature decrease.

Hence, option $C$ is the correct answer.

$P_{2} V_{2}=P_{1} V_{1}$

$P_{2}=P_{1} \frac{V_{1}}{V_{2}}=P_{1} \frac{V_{1}}{V_{1} / 2}=2 P_{1}$

For gas in $\mathrm{B}$, when compression is adiabatic,

$P_{2}^{\prime} V_{2}^{\prime}=P_{1} V_{1}^{\gamma}$

$P_{2}^{\prime}=P_{1}\left(\frac{V_{1}}{V_{2}^{\prime}}\right)^{\gamma}=P_{1}\left(\frac{V_{1}}{V_{1} / 2}\right)^{\gamma}=2^{\gamma} P_{1}$

$\therefore \frac{P_{2}^{\prime}}{P_{2}}=\frac{2^{\gamma} P_{1}}{2 P_{1}}=2^{\gamma-1}$

$\Rightarrow \mathrm{PV}=\mathrm{P}_{2} \times 4 \mathrm{V} \quad \because \mathrm{P}_{2}=\mathrm{P} / 4$

In adiabatic process

$\mathrm{P}_{2} \mathrm{V}_{2}^{\gamma}=\mathrm{P}_{3} \mathrm{V}_{3}^{\gamma} \Rightarrow \frac{\mathrm{P}}{4} \times(4 \mathrm{V})^{15}=\mathrm{P}_{3} \mathrm{V}^{15} \Rightarrow \mathrm{P}_{3}=2 \mathrm{P}$

$146 \times 10^{3}=\frac{10^{3} \times 8.3 \times 7}{\gamma-1}$

$\gamma-1=\frac{8.3 \times 7}{146}=\frac{58.1}{146}$

$\gamma=\frac{58}{146}+1$

$\gamma=\frac{58+146}{146}$

$=\frac{204}{146}$

$=\frac{7}{5}$ diatomic nature.

$T_{A}>T_{B}$

$\mathrm{P}_{1} \mathrm{V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{V}_{2}^{\gamma}$

or $\quad \mathrm{T} \mathrm{P}^{(1-\gamma) / \gamma}=$ constant

$\mathrm{T}_{1} \mathrm{P}_{1}^{(1-\mathrm{\gamma}) / \gamma}=\mathrm{T}_{2} \mathrm{P}_{2}^{(1-\gamma) / \gamma}$

$\mathrm{T}_{2}=\mathrm{T}_{1}\left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right)^{(\mathrm{1}-\mathrm{y}) / \gamma}=\mathrm{T}_{1}(2)^{\frac{-0.4}{1.4}}=\frac{300}{(4)^{1 / 7}}$

$\Rightarrow \mathrm{T}_{2}=246 \mathrm{K}=-27^{\circ} \mathrm{C}$

And similarly, $T \times V^{\gamma-1}=l$ where $l$ is a constant.

Therefore, the decrease in volume is associated with increase in temperature as well as increase in pressure.

$P \propto T^2$

$P T^{-2}=$ constant $\quad$ compare with $P T^{\left(\frac{\gamma}{1-\gamma}\right)}=$ constant

$\frac{C_P}{C_V}=\gamma=2$

$T V^{1-1}=\text { constant }$

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

$T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}$

$T_2=(273 \,K )\left[\frac{V_1}{V / 9}\right]^{1.5-1}$

$T_2=(273 \,K ) \times 3$

$=819 \,K$

$=546^{\circ} C$

Adiabatic elasticity $=\gamma P$

$=\frac{5}{3} \times 3 \times 10^5=5 \times 10^5 \,N / m ^2$

$T V^{\gamma-1}=K$

$\log T+(\gamma-1) \log V=0$

$\log T=-(\gamma-1) \log V$

$y=-(\gamma-1) x$

$\frac{y}{x}=-(\gamma-1)=\text { slope }=\frac{2-4}{4-1}$

$\Rightarrow-(\gamma-1)=-\frac{2}{3}$

$\gamma=\frac{5}{3}$

$\therefore \text { Monoatomic.}$

$P_1 V_1=P_2 V_2$   [for isothermal]

$P V=P \times 2 V$

$\frac{P}{2}=P^{\prime}$

$P_1 V_1^\gamma=P_2 V_2^\gamma$   [for adiabatic]

$\frac{P}{2} \times(2 V)^{5 / 3}=P_2(V)^{5 / 3}$    $[\gamma$ for neon $=5 / 3]$

$P=P_2 \cdot(2)^{-2 / 3}$

Fractional decrease $=\frac{P_2-P}{P_2}=\frac{P_2-P_2 \cdot(2)^{-2 / 3}}{P_2}=1-2^{-2 / 3}$

Also$PV = \mu RT$==>$\frac{k}{{{V^{1/3}}}}.V = \mu RT$==> ${V^{2/3}} = \frac{{\mu RT}}{k}$

Hence ${\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{2/3}} = \frac{{{T_1}}}{{{T_2}}}$==> ${\left( {\frac{V}{{27V}}} \right)^{2/3}} = \frac{T}{{{T_2}}}$==> ${T_2} = 9\,T$

According to question $T \propto {V^{ - \frac{1}{2}}}$

Hence $1 - \gamma = - \frac{1}{2} \Rightarrow \gamma \frac{3}{2} = 1.5$

Since $V = k{T^{2/3}}$ ==> $dV = \frac{2}{3}K{T^{ - 1/3}}dT$

Eliminating $K,$ we find $\frac{{dV}}{V} = \frac{2}{3}\frac{{dT}}{T}$

Hence$W = \int_{{T_1}}^{{T_2}} {\,\frac{2}{3}\frac{{RT}}{T}dT} = \frac{2}{3}R({T_2} - {T_1}) = \frac{2}{3}R(30) = 20\,R$

proportional to $V.$

Adiabatic process

$\mathrm{pV}^{\gamma}=$ constant

$\mathrm{p}\left(\frac{\mathrm{M}}{\rho}\right)^{\gamma}=\mathrm{constant}$

$\mathrm{p} \cdot \mathrm{M}^{\gamma} \cdot \rho^{-\gamma}=$ constant $\quad\left(\mathrm{M}^{\gamma} \text { is constant }\right)$

$\mathrm{p} \cdot \rho^{-\gamma}=$ constant

in the given process

$\mathrm{P} \alpha \mathrm{V}$

$\mathrm{PV}^{-1}=\mathrm{const}$

$n=-1$

$\Rightarrow \mathrm{C}=\frac{3}{2} \mathrm{R}+\frac{\mathrm{R}}{1-(-1)}=2 \mathrm{R}$

$W=\frac{n R \Delta T}{1-x}$

$=5 \times 10^{5} \int_{1}^{2} \mathrm{V}^{2} \mathrm{d} \mathrm{v}$

$=5 \times 10^{5}\left[\frac{V^{3}}{3}\right]_{1}^{2}$

$=\frac{5}{3} \times\left[2^{3}-1^{3}\right] \times 10^{5} \mathrm{J}$

$=1018 \mathrm{MJ}$

$\left(\frac{\mathrm{nRT}}{\mathrm{V}}\right)^{2} \mathrm{V}=\mathrm{constant}$

$\frac{\mathrm{T}^{2}}{\mathrm{V}}=\mathrm{constant}$

$\frac{\mathrm{T}_{1}^{2}}{\mathrm{V}_{1}}=\frac{\mathrm{T}_{2}^{2}}{\mathrm{V}_{2}} \quad \Rightarrow \quad \mathrm{T}_{2}^{2}=\frac{\mathrm{V}_{2}}{\mathrm{V}_{1}} \mathrm{T}_{1}^{2}=2 \mathrm{T}_{0}^{2}$

$\Rightarrow \mathrm{T}_{2}=\sqrt{2} \mathrm{T}_{0}$

$\mathrm{P}=0.40 \mathrm{V}^{-2.5}$

$\frac{\mathrm{dP}}{\mathrm{dV}}=0.40\left[-2.5 \mathrm{V}^{-3.5}\right]$

at $\quad V=1$

$\left(\frac{\mathrm{dP}}{\mathrm{dV}}\right)_{\mathrm{V}=1}=-1$

$P \propto V^{-2}$

$P V^2=$ constant $\quad$ Compare with $P V^N=$ constant then $N=2$

$W=\mu\left(\frac{R}{1-N}\right) \Delta T$

$W=\frac{\mu R}{1-N}\left(T_2-T_1\right)$

$=\frac{2 \times R(400-300)}{(1-2)}$

$=-200 R$

$P V^{1.3}=K$

$W=\frac{P_2 V_2-P_1 V_1}{1-N}$

$\because N > 1$, so $W$ is negative.

Heat supplied by surrounding heat goes to do work.

$\therefore$ Down when expands.

Work done $\Delta W = -50 J$         [As process is anticlockwise]

By first law of thermodynamics

==> $\Delta U = \Delta Q - \Delta W = 84 - ( - \,50)\, = 134J$

Process $BC$ is isothermal $\therefore$  ${W_{BC}} = R{T_2}.\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right)$ 

Process $CA$ is isobaric 

$\therefore  {W_{CA}} = - \,P\Delta V$$ = - \,R\Delta T$$ = - \,R({T_1} - {T_2})$$ = R({T_2} - {T_1})$

(Negative sign is taken because of compression)

= $(2V - V) \times (2P - P) = PV$

 $\because  \Delta {W_A} > \Delta {W_B}$ ==> $\Delta {Q_A} > \Delta {Q_B}$

Treat the circle as an ellipse of area $ = \frac{\pi }{4}({P_2} - {P_1})\,({V_2} - {V_1})$

$\Rightarrow \Delta Q = \frac{\pi }{4}\{ (150 - 50) \times {10^3}\} = \frac{\pi }{2}J$

$ = \frac{1}{2}(1 \times {10^5} + 5 \times {10^5}) \times (5 - 1) = 12 \times {10^5}J$

For path $iaf : \Delta U = \Delta Q - \Delta W = 50 - 20 = 30J$. 

For path $fi  :  \Delta U = - \,30\,J\,{\rm{and\, }}\Delta W = - 13J$ 

==> $\Delta Q = - \,30 - 13 = - \,43\;J$.

$ = \frac{1}{2}({P_A} + {P_B})\,({V_B} - {V_A})$

$⇒$ $\Delta  U =  \Delta  Q -  \Delta  W = 40 -30 = 10 J.$

$\Delta U = \frac{f}{2}\mu R\Delta T = \frac{f}{2}({P_f}{V_f} - {P_i}{V_i})$

$ = \frac{3}{2}(2{P_0} \times 2{V_0} - {P_0} \times {V_0}) = \frac{9}{2}{P_0}{V_0}$

Work done in process $A \to B$ is equal to the Area covered by the graph with volume axis i.e.,

${W_{A \to B}} = \frac{1}{2}({P_0} + 2{P_0}) \times (2{V_0} - {V_0}) = \frac{3}{2}{P_0}{V_0}$

Hence, $\Delta Q = \Delta U + \Delta W$$ = \frac{9}{2}{P_0}{V_0} + \frac{3}{2}{P_0}{V_0} = 6{P_0}{V_0}$

$\Delta Q = $heat supplied to diatomic gas at constant $P$

$\therefore \;\frac{{\Delta W}}{{\Delta Q}} = \frac{{RdT}}{{\frac{7}{2}RdT}} = \frac{2}{7}$

==> ${W_{ABCD}} = - \,{P_0}{V_0} + 0 + 4{P_0}{V_0} = 3{P_0}{V_0}$

$\therefore $ $\Delta Q = n{C_P}\Delta T$ and $\Delta U = n{C_V}\Delta T$

Also $\Delta W = \Delta Q - \Delta U = \mu \,({C_P} - {C_V})\,\Delta T$

$\therefore \,\,\,\,\,\,\frac{{\Delta Q}}{{\Delta W}} = \frac{{n{C_P}\Delta T}}{{n\,({C_P} - {C_V})\,\Delta T}}$$ = \frac{{{C_P}}}{{{C_P} - {C_V}}} = \frac{1}{{1 - \frac{{{C_V}}}{{{C_P}}}}}$

$\frac{{{C_V}}}{{{C_P}}} = \frac{3}{5}$ for helium gas. Hence $\frac{{\Delta Q}}{{\Delta W}} = \frac{1}{{1 - 3/5}} = \frac{5}{2}$

$\Delta U = \mu {C_V}\Delta T = \frac{5}{2}\mu R\Delta T$ $\left( {{C_V} = \frac{5}{2}R} \right)$

and $\Delta W = \Delta Q - \Delta U = \mu R\Delta T$

==> $\Delta Q\,:\,\Delta U\,\,:\,\,\Delta W = 7\,\,:\,\,5\,\,:\,\,2$

Initial internal energy of the gas is ${U_1} = N\,\left( {\frac{5}{2}R} \right)\,T$

Since $n$ moles get dissociated into atoms, therefore, after heating, vessel contains $(N - n)$ moles of diatomic gas and $2n$ moles of a mono-atomic gas. Hence the internal energy for the gas, after heating, will be equal to

${U_2} = (N - n)\left( {\frac{5}{2}R} \right)\,T + 2n\,\left( {\frac{3}{2}R} \right)\,T$$ = \frac{5}{2}\,NRT + \,\frac{1}{2}nRT$

Hence, the heat supplied = increase in internal energy

$ = \,({U_2} - {U_1})\, = \,\frac{1}{2}\,nRT$

In process $BC,$ the temperature increases linearly; hence the internal energy of the gas also increases.

$W=P \Delta V$

Given that

$\mathrm{P}=$ Constant

And volume is doubled from initial position.

$W=P(2 V-V)$

$=P V$

But, by ldeal gas equation $P V=n R T$

For $1$ mole gas, $P V=R T$

$\Rightarrow W=R T$

$w=\int P d V$

Work done by $x$ is against vacuum, hence $P=0$

$w=\int P d V=\int 0 d V=0$

Hence no work is done by or done on $x$.

Work $ = P (8V -V) = 7PV$ [isobaric process]

$[\mathrm{as} W=p \Delta V \text { and } \Delta V=0, \text { so } W=0]$

$\Delta U=m C \Delta T$

$=100 \times 10^{-3} \times 4184 \times(50-30) \approx 8.4 \mathrm{kJ}$

$5=0+10(2-1)+0+\mathrm{W}_{\mathrm{CA}}$

$\mathrm{W}_{\mathrm{CA}}=5-10$

$=-5$ Joule

$\Rightarrow \mathrm{Q}_{\mathrm{ABCA}}=\mathrm{W}_{\mathrm{ABCA}}$

$\Rightarrow \mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}+\mathrm{Q}_{\mathrm{B} \rightarrow \mathrm{C}}+\mathrm{Q}_{\mathrm{C} \rightarrow \mathrm{A}}$

$=\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}+\mathrm{W}_{\mathrm{B} \rightarrow \mathrm{C}}+\mathrm{W}_{\mathrm{C} \rightarrow \mathrm{A}}$   $.(1)$

Given $Q_{B \rightarrow C}+Q_{C \rightarrow A}=W_{A \rightarrow B}+W_{B \rightarrow C}$  $(2)$

Subtracting $( 2)$ from $( 1)$

$\mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}}=\mathrm{W}_{\mathrm{C} \rightarrow \mathrm{A}}=0$

(as process $\mathrm{C} \rightarrow \mathrm{A}$ is isochoric)

process $A \rightarrow B$ is adiabatic.

$5 \mathrm{R}=2 \mathrm{C}_{\mathrm{p}}=2\left[\mathrm{C}_{\mathrm{v}}+\mathrm{R}\right]$

$C_{v}=\frac{3 R}{2}$

$\therefore(1)$

$f=\frac{\Delta \mathrm{U}}{(\Delta \mathrm{Q})_{\mathrm{P}}}=\frac{(\Delta \mathrm{Q})_{\mathrm{V}}}{(\Delta \mathrm{Q})_{\mathrm{P}}}=\frac{\mu \mathrm{C}_{\mathrm{V}} \Delta \mathrm{T}}{\mu \mathrm{C}_{\mathrm{P}} \Delta \mathrm{T}}=\frac{1}{\gamma}$

For diatomic gas $\gamma=\frac{7}{5} \Rightarrow f=\frac{5}{7}$

$dQ=0+600$

$dQ=600\,J$

$=1 \times 8.31 \times 100=831$

The process $\mathrm{BC}$ is an isochoric process since the volume is constant

The process $\mathrm{CA}$ is isothermal since, temperature remains constant.

The only $P-v$ graph that matches the above criteria is option $\mathrm{C}$

$\mathrm{d} \mathrm{W}=168 \mathrm{J}$

$\mathrm{dQ}=\mathrm{mL}=\mathrm{dU}+\mathrm{d} \mathrm{W}$

or $1 \times 2240=d U+168$

$\mathrm{dU}=2072 \mathrm{J}$

$=\mu \frac{1}{2} \mathrm{R}\left(2 \mathrm{T}_{0}-\mathrm{T}_{0}\right)$

$=\frac{5}{2} \mathrm{P}_{0} \mathrm{V}_{0}$

$P=$ const

$\mathrm{U}_{\mathrm{C}}-\mathrm{U}_{\mathrm{B}}=0$

$\mathrm{T} \uparrow$

$\mathrm{U}_{\mathrm{B}}-\mathrm{U}_{\mathrm{A}}>0$

$\mathrm{C} \rightarrow \mathrm{D}$

$\mathrm{U}_{\mathrm{C}}-\mathrm{U}_{\mathrm{P}}>0$

$\Rightarrow \Delta \mathrm{Q}_{\mathrm{ABC}}-\mathrm{W}_{\mathrm{ABC}}=\Delta \mathrm{Q}_{\mathrm{ADC}}-\mathrm{W}_{\mathrm{ADC}}$

$\Rightarrow 90-30=\Delta Q_{A D C}-20$

$\Rightarrow \Delta \mathrm{Q}_{\mathrm{ADC}}=80 \mathrm{J}$

$(600+200)=\Delta \mathrm{U}_{\mathrm{AC}}+\left[8 \times 10^{4}(5-2) \times 10^{-3}\right]$

$\Delta \mathrm{U}_{\mathrm{AC}}=800-240$

$=560$ Jule.

$\frac{16 \mathrm{P}_{0}}{\mathrm{P}_{0}}=\left(\frac{2 \mathrm{V}_{0}}{\mathrm{V}_{0}}\right)^{\mathrm{x}}$

$x=4$

$\mathrm{w}=\frac{\mathrm{P}_{2} \mathrm{V}_{2}-\mathrm{P}_{1} \mathrm{V}_{1}}{1-\mathrm{x}}=\frac{2 \mathrm{P}_{0} \mathrm{V}_{0}-16 \mathrm{P}_{0} \mathrm{V}_{0}}{1-4}$

$=\frac{14}{3} \mathrm{P}_{0} \mathrm{V}_{0}$

Change in internal energy $=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$

$\therefore$ Fraction $=\frac{C_{v}}{C_{p}}=\frac{1}{\gamma}$

For a non$-$linear triatomic gas $\gamma=\frac{4}{3}$

fraction $=\frac{3}{4}$

$\left(\frac{P}{T}\right)_{A}=\frac{P_{0}}{T_{0}}$ and $\left(\frac{P}{T}\right)_{B}$

$=\frac{3}{2}\left(\frac{P_{0}}{T_{0}}\right)\left(\frac{P}{T}\right)_{B}=\frac{3}{2}\left(\frac{P}{T}\right)_{A}$

$\therefore \quad \rho_{B}=\frac{3}{2} \rho_{A}=\frac{3}{2} \rho_{0}$

$\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}=1 \times \frac{3}{2} \mathrm{R} \times\left(2 \mathrm{T}_{0}-\mathrm{T}_{0}\right)=\frac{3}{2} \mathrm{RT}_{0}$

Work $=\frac{\mu R \Delta T}{n-1}=\frac{1 \times R \times\left(2 T_{0}-T_{0}\right)}{2} \Rightarrow\left(\frac{R T_{0}}{2}\right)$

$\mathrm{Q}=\frac{3}{2} \mathrm{RT}_{0}+\frac{\mathrm{RT}_{0}}{2}$

$=2 \mathrm{RT}_{0}$

$\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=\mathrm{n} \mathrm{R} \Delta \mathrm{T}$

But $Q=n C_{p} \Delta T=n \frac{\gamma R}{r-1}$

$\Rightarrow \quad W=Q\left(\frac{\gamma-1}{r}\right)=100\left(\frac{1.4-1}{1.4}\right)=28.57 \mathrm{J}$

$\Delta U =$ Zero

$\mathrm{dQ}=\mathrm{d} \mathrm{W}$

$\Rightarrow \mathrm{d} \mathrm{Q}=\frac{1}{2} \times 200 \times 100 \times 10^{-6} \times 10^{3} \mathrm{J}=10 \mathrm{Joule}$

also $dQ$ $=2.4 \mathrm{cal}$

$\Rightarrow \mathrm{J}=\frac{10}{2.4}=4.17 \mathrm{J} / \mathrm{cal}$

$\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$

Here, $\Delta \mathrm{U}$ is same for all the three processes as it depends only on initial and final states.

$\text { But } \quad \Delta \mathrm{W}_{1}=+\mathrm{ve}, \Delta \mathrm{W}_{\mathrm{II}}=0$

${\text { and }}  {\Delta \mathrm{W}_{\mathrm{III}}=-\mathrm{ve}} $

${\text { or }}  {\Delta \mathrm{Q}_{1}>\Delta \mathrm{Q}_{\mathrm{II}}}$

$d W=-8 J$

$V_{i}=30 J$

$d Q=d V+d W$

$d Q=\left(V_{f}-V_{i}\right)+d W$

$-20=\left(V_{f}-30\right)-8$

$-12=V_{f}-30$

$V_{f}=18$

$\therefore$ Final internal energy $=18 J$

$\mathrm{W}=\mathrm{W}_{\mathrm{ABCD}}+\mathrm{W}_{\mathrm{DEAD}}$

$\mathrm{W}=1-\frac{1}{2}=\frac{1}{2} \mathrm{J}$

$-50=-20+\left(U_{t}-U_{1}\right)$

$\mathrm{U}_{\mathrm{f}}-(-30)=-30$

$\mathrm{U}_{\mathrm{f}}=-60 \mathrm{Cal}$

Cyclic process is anticlockwise then

Work done $=-$ (Area of $P-V$ graph $)$

$W=-\pi R_1 R_2$

$=-\pi\left(\frac{3 P_0-P_0}{2}\right) \times\left(\frac{3 V_0-V_0}{2}\right)$

$=\frac{-22}{7} P_0 V_0$

Area under graph and $V$ axis $=$ work done

$=\frac{1}{2} \times(30+10) \times 10^3 \times(25-10) \times 10^{-6}$

$=0.3 \,J$

When taken through $A B C[\triangle U+$ work $=$ heat absorbed $]$

Heat absorbed $=$ area under graph $+\Delta U=8$

$\Delta U=8-\frac{10 \times 200}{1000}=6$

when taken directly to $C$

$W+\Delta U=Q$

${\left[\frac{10 \times 200}{1000}+\frac{1}{2} \times \frac{2000}{1000}\right]+6=Q \Rightarrow Q=9 \,J }$

In process $B C$ (isochoric process) where $\Delta T$ is (+)ive.

So $\Delta U=n C_v \Delta T$

$\because \Delta T$ is positive $\Rightarrow U$ increases

==> $Q = \left( {\frac{{{T_1}}}{{{T_1} - {T_2}}}} \right)\,W$

$ = \frac{{600}}{{(600 - 300)}} \times 800 =1600 \,J$

${T_1} = 400\,K = 127^\circ C$

Then $Q_{2}=3 W$

Also, $Q_{1}=Q_{2}+W$

$\Rightarrow Q_{1}=4 W$

Efficiency of cycle=work done/heat taken $=\frac{W}{4 W}=\frac{1}{4}=0.25$