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\% $

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

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 $IV \to Isobaric;\,S \to B$

$VT=$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, $Q=n C \Delta T$

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

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

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

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

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

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

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

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.

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.

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

$T_3 > T_2 > T_1$

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$.

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

==> ${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$

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

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

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

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