Question

One mole of an ideal monoatomic gas is taken round cyclic process ABCA as shown in figure. Calculate :
(a) The work done by the gas.
(b) The heat rejected by the gas in the path CA, the heat absorbed bythe gas in the path AB.
(c) The net heat absorbed by the gas in the path BC.
(d) The maximum temperature attained by the gas during the cycle.

Answer 1

Clockwise cyclic process

$a)$ Work done by gas

W = area($△$ABC)

$=\frac{1}{2}(2V_o-V_o)(3P_o-P_o)=P_o{V}_o$

$b)$ n = 1 monoatomic gas

$C_V=\frac{3}{2}R\Rightarrow \frac{C_V}{R}=\frac{3}{2}{C}_P=\frac{5}{2}R\Rightarrow \frac{C_P}{R}=\frac{5}{2}$

Heat rejected by gas in the path CA

CA $\to$ It is isochoric

$d{\theta }_{CA}=\frac{5}{2}(P_o{V}_o-2P_o{V}_o)=\frac{-5}{2}(P_o{V}_o-2P_o{V}_o)$

Heat absorbed by the gas in path AB

AB $\to$ isochoric

$\therefore d{\theta }_{AB}=C_{V}dT=C_V(T_{final}-T_{initial})=C_V(\frac{P_f{V}_f}{R}-\frac{P_i{V}_i}{R})=\frac{C_V}{R}(P_f{V}_f-P_i{V}_i)=\frac{3}{2}(P_f{V}_f-P_i{V}_i)=\frac{3}{2}(3P_o{V}_o-P_o{V}_o)=3P_o{V}_o$

$c)$ Total heat absorbed during the process

$d\theta =d{\theta }_{AB}+d{\theta }_{BC}+d{\theta }_{CA}=3P_o{V}_o+d{\theta }_{BC}+\left(\frac{-5}{2}{P}_o{V}_o\right)d\theta =\frac{P_o{V}_o}{2}+d{\theta }_{BC}$

Change in internal energy

$dU=0\Rightarrow d\theta =dW(d\theta =dU+dW)\frac{P_o{V}_o}{2}+d{\theta }_{BC}=P_o{V}_{o}d{\theta }_{BC}=\frac{P_o{V}_o}{2}$

$d)$ PV equation for the process BC is

$P=-mV+C$ (max temperature will be between B and C)

here

$m=\frac{2P_o}{V_o},C=5P_o\therefore P=-\left(\frac{2P_o}{V_o}\right)V+5P_{o}PV=-\left(\frac{2P_o}{V_o}\right)V^2+5P_{o}VRT=-\left(\frac{2P_o}{V_o}\right)V^2+5P_{o}V[PV=RT;forn=1]T=\frac{1}{R}[5P_{o}V-\left(\frac{2P_o}{V_o}\right)V^2]\frac{dT}{dV}=T\Rightarrow 5P_o-\frac{4P_o}{V_o}V=0\Rightarrow V=\frac{5V_o}{4}at,V=\frac{5V_o}{4};T=maximum{T}_{max}=\frac{1}{R}[5P_o\left(\frac{5V_o}{4}\right)-\left(\frac{2P_o}{V_o}\right){\left(\frac{5V_o}{4}\right)}^2]T_{max}=\frac{25}{8}\frac{P_o{V}_o}{R}$