8.0 Refrigerator

Refrigerator is a device which takes heat from a cold body, work is done on it and the work done together with the heat absorbed is rejected to the source.An ideal refrigerator can be regarded as carnots ideal heat engine working in the reverse direction.

$\left( \beta \right)$ of a refrigerator is defined as the ratio of quantity of heat removed per cycle ${\left( {{Q_2}} \right)}$ to the work done on the working substance per cycle to remove this heat. Thus,$$\beta = \frac{{{Q_2}}}{W} = \frac{{{Q_2}}}{{{Q_1} - {Q_2}}}$$$$\beta = \frac{{{T_2}}}{{{T_1} - {T_2}}} = \frac{{1 - \eta }}{\eta }$$ Here, $\eta $ is the efficiency of carnots cycle.

Calculate the least amount of work that must be done to freeze one gram of water at ${0^o}C$ by means of a refrigerator. Temperature of surroundings is ${27^o}C$. How much heat is passed on the surroundings in this pricess? Latent heat of usion $L = 80\,cal/g$.

$$\begin{equation} \begin{aligned} {Q_2} = mL = 1 \times 80 = 80\,cal \\ {T_2} = {0^O}C = 273K \\ and\quad {T_1} = {27^O}C = 300K \\ \frac{{{Q_2}}}{W} = \frac{{{T_2}}}{{{T_1} - {T_2}}} \\ W = \frac{{{Q_2}\left( {{T_1} - {T_2}} \right)}}{{{T_2}}} \\ = \frac{{80\left( {300 - 273} \right)}}{{273}} = 7.91\,cal \\ \\ {Q_1} = {Q_2} + W \\ = \left( {80 + 7.91} \right) = 87.91\,cal \\\end{aligned} \end{equation} $$

The density versus pressure graph of one mole of an ideal monoatomic gas undergoing a cyclic process is shown in figure. The molecular mass of the gas is $M$.

(a) Find the work done in each process.

(b) Find heat rejected by gas in one complete cycle.

(c) Find the efficiency of the cycle

(a) As $n = 1,\;m = M$

Process $AB$ : $\rho \propto P$, It is an isothermal process ($T$ = constant ), because $\rho = \frac{{PM}}{{RT}}$.$$\begin{equation} \begin{aligned} {W_{AB}} = R{T_A}\ln \left( {\frac{{{P_A}}}{{{P_B}}}} \right) = R{T_A}\ln \left( {\frac{1}{2}} \right) \\ = - \frac{{{P_0}M}}{{{\rho _0}}}\ln \left( 2 \right) \\ \Delta {U_{AB}} = 0 \\ and\quad {Q_{AB}} = {W_{AB}} = - \frac{{{P_0}M}}{{{\rho _0}}}\ln \left( 2 \right) \\\end{aligned} \end{equation} $$Process $BC$ is an isobaric process ($P$ = constant) $$\begin{equation} \begin{aligned} {W_{BC}} = {P_B}\left( {{V_C} - {V_B}} \right) = 2{P_0}\left( {\frac{M}{{{\rho _C}}} - \frac{M}{{{\rho _B}}}} \right) = \frac{{2{P_0}M}}{{2{\rho _0}}} = \frac{{{P_0}M}}{{{\rho _0}}} \\ \Delta {U_{BC}} = {C_V}\Delta T \\ = \left( {\frac{3}{2}R} \right)\left[ {\frac{{2{P_0}M}}{{{\rho _0}R}} - \frac{{2{P_0}M}}{{2{\rho _0}R}}} \right] = \frac{{3{P_0}M}}{{2{\rho _0}}} \\ {Q_{BC}} = {W_{BC}} + \Delta {U_{BC}} = \frac{{5{P_0}M}}{{2{\rho _0}}} \\\end{aligned} \end{equation} $$Process $CA$ : As $\rho $= constant $\therefore \;V$.

So, it is an isochoric process.$$\begin{equation} \begin{aligned} {W_{CA}} = 0 \\ \Delta {U_{CA}} = {C_V}\Delta T \\ = \left( {\frac{3}{2}R} \right)\left( {{T_A} - {T_C}} \right) \\ = \left( {\frac{3}{2}R} \right)\left[ {\frac{{{P_0}M}}{{{\rho _0}R}} - \frac{{2{P_0}M}}{{{\rho _0}R}}} \right] = - \frac{{3{P_0}M}}{{2{\rho _0}}} \\ {Q_{CA}} = \Delta {U_{CA}} = - \frac{{3{P_0}M}}{{2{\rho _0}}} \\\end{aligned} \end{equation} $$(b) Heat rejected by the gas = $$\begin{equation} \begin{aligned} \left| {{Q_{AB}}} \right| + \left| {{Q_{CA}}} \right| \\ = \frac{{{P_0}M}}{{{\rho _0}}}\left[ {\frac{3}{2} + \ln \left( 2 \right)} \right] \\\end{aligned} \end{equation} $$(c) Efficiency of the cycle (in fraction )$$\begin{equation} \begin{aligned} \eta = \frac{Total work done}{Heat supplied} = \frac{{{W_{Total}}}}{{{Q_{ + ve}}}} \\ = \frac{{\frac{{{P_0}M}}{{{\rho _0}}}\left[ {1 - \ln \left( 2 \right)} \right]}}{{\frac{5}{2}\left( {\frac{{{P_0}M}}{{{\rho _0}}}} \right)}} \\ = \frac{2}{5}\left[ {1 - \ln \left( 2 \right)} \right] \\\end{aligned} \end{equation} $$