Question

One mole of an ideal monoatomic gas is taken round the cyclic process ABCA as shown in the figure, calculate the heat rejected by the gas in the path CA and the heat absorbed by the gas in the path AB.

Answer 1

Heat rejected by the gas in CA $\Delta Q=n{C}_p\Delta T$

we have $n=1$

and we know $C_p=\frac{5R}{2}$ for monoatomic gas

$PV=nRT$

$T_i=\frac{P_o\times 2V_o}{R}$ $T_f=\frac{P_o{V}_o}{R}$

$\Delta T$ = $\frac{P_o{V}_0}{R}-\frac{2P_o{V}_o}{R}$

putting the values into the equation

$Q=\frac{5R}{2}\times \left(\frac{-P_o{V}_o}{R}\right)=-\frac{5P_o{V}_o}{2}$ [negative sign shows that the heat was given out in the process]

Hence heat rejected is $\frac{5P_o{V}_o}{2}$

Heat absorbed in AB is

$\Delta Q=n{C}_v\Delta T$

we know $C_v=\frac{3R}{2}$ for monoatomic gas

$T_i=\frac{P_o{V}_o}{R}$ $T_f=\frac{3P_o\times V_o}{R}$

putting the values into the equation

$Q=\frac{3R}{2}\times (\frac{3P_0{V}_o}{R}-\frac{P_0{V}_0}{R})=3P_o{V}_o$