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.