Question

One mole of an ideal monatomic gas is taken around the cyclic ABCA as shown in the figure. Calculate
$(A)$ The work done by the gas
$(B)$ The heat rejected by the gas in the path CA and the heat absorbed by the gas in 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

The system starts and returns to the same thermodynamic state in a cyclic process.
Internal energy is a function of the state. The internal energy of a cyclic process remains unchanged because the system returns to its initial state.
The work done by the system is equal to the heat added to it. An adiabatic change that does no work is known as a free expansion.

Complete step-by-step solution:
It is a clockwise cyclic process.
$(A)$Work done by gas
$W =$ area $\left( {\Delta ABC} \right)$
So,
$= \dfrac{1}{2}\left( {2{V_0} - {V_0}} \right)\left( {3{P_0} - {P_0}} \right)$
$= {P_0}{V_0}$

$(B)$ The heat rejected by the gas in the path CA and the heat absorbed by the gas in path AB
$n = 1$ mole
so,
${C_V} = \dfrac{3}{2}R \Rightarrow \dfrac{{{C_v}}}{R} = \dfrac{3}{2}$
${C_P} = \dfrac{5}{2}R \Rightarrow \dfrac{{{C_P}}}{R} = \dfrac{5}{2}$
Now, heat rejected by the gas in path CA, it is isochoric
$d{\theta _{CA}} = \dfrac{5}{2}\left( {{P_0}{V_0} - 2{P_0}{V_0}} \right) = \dfrac{{ - 5}}{2}\left( {{P_0}{V_0} - 2{P_0}{V_0}} \right)$
Now, for the heat absorbed by the gas in path AB, it is isochoric.
$d{\theta _{AB}} = {C_V}dT$
$\Rightarrow d{\theta _{AB}} = {C_V}\left( {{T_{final}} - {T_{initial}}} \right)$
$\Rightarrow d{\theta _{AB}} = {C_V}\left( {\dfrac{{{P_f}{V_f}}}{R} - \dfrac{{{P_i}{V_i}}}{R}} \right)$
$\Rightarrow d{\theta _{AB}} = \dfrac{{{C_V}}}{R}\left( {{P_f}{V_f} - {P_i}{V_i}} \right)$
$\Rightarrow d{\theta _{AB}} = \dfrac{3}{2}\left( {{P_f}{V_f} - {P_i}{V_i}} \right)$
$\Rightarrow d{\theta _{AB}} = \dfrac{3}{2}\left( {3{P_0}{V_0} - {P_0}{V_0}} \right) = 3{P_0}{V_0}$
$(C)$Total heat absorbed during the process
$d\theta = d{\theta _{AB}} + d{\theta _{BC}} + d{\theta _{CA}}$
$\Rightarrow d\theta = 3{P_0}{V_0} + d{\theta _{BC}} + \left( {\dfrac{{ - 5}}{2}{P_0}{V_0}} \right)$
$\Rightarrow d\theta = \dfrac{{{P_0}{V_0}}}{2} + d{\theta _{BC}}$
Change in internal energy $dU = 0$
$d\theta = dU + dW$
$\Rightarrow d\theta = dW$
$\Rightarrow \dfrac{{{P_0}{V_0}}}{2} + d{\theta _{BC}} = {P_0}{V_0}$
$\Rightarrow d{\theta _{BC}} = \dfrac{{{P_0}{V_0}}}{2}$
$(D)$ $PV$ equation for the process BC is
$P = - mV + C$
(max temperature will be between B and C)
Here,
$m = \dfrac{{2{P_0}}}{{{V_0}}},C = 5{P_0}$
$\therefore P = - \left( {\dfrac{{2{P_0}}}{{{V_0}}}} \right)V + 5{P_{{0_{}}}}$
$PV = - \left( {\dfrac{{2{P_0}}}{{{V_0}}}} \right){V^2} + 5{P_0}V$
$RT = - \left( {\dfrac{{2{P_0}}}{{{V_0}}}} \right){V^2} + 5{P_0}V$
$T = \dfrac{1}{R}\left[ {5{P_0}V - \left( {\dfrac{{2{P_0}}}{{{V_0}}}} \right){V^2}} \right]$
$\dfrac{{dT}}{{dV}} = T$
$\Rightarrow 5{P_0} - \dfrac{{4{P_0}}}{{{V_0}}}V = 0$
$\Rightarrow V = \dfrac{{5{V_0}}}{4}$
$at,V = \dfrac{{5{V_0}}}{4};$
$T = \max imum$
${T_{\max }} = \dfrac{1}{R}\left[ {5{P_0}\left( {\dfrac{{5{V_0}}}{4}} \right) - \left( {\dfrac{{2{P_0}}}{{{V_0}}}} \right){{\left( {\dfrac{{5{V_0}}}{4}} \right)}^2}} \right]$
${T_{\max }} = \dfrac{{25}}{8}\dfrac{{{P_0}{V_0}}}{R}$

Note: The system is functional if the cycle rotates clockwise. If the clockwise cycle is followed, work on the system is completed every cycle.
Because the system returns to its initial condition in a cyclic process, the change in internal energy must be zero. As a result, the net heat delivered to the system must equal the net work done by the system, according to the first law of thermodynamics.
Because gas compresses around its environment when it expands, the work done by the gas is equal to the force it exerts on it multiplied by the distance traveled. The magnitude of work done can also be determined from the graph by calculating the change in pressure and volume.