Answer
a)We draw the $p-V$ diagram.
b) $2462.8J$ work is done on the gas in the process $ca$.
Work Step by Step
b) For the cycle, total change in internal energy ($\Delta U$) $=0$
(Internal energy is a state function and we return to the same state after a cycle)
Heat output in the cycle $\Delta Q=-800J$
(Heat flowing out of system is negative)
By first law of thermodynamics, $\Delta Q=\Delta U+\Delta W=0+\Delta W=\Delta W=-800J$
($\Delta W$ is the work done in the cycle)
Pressure at $a=P_a$
Pressure at $b=P_b$
$P_a=P_b=P$ ($ab$ is a constant pressure process)
Volume of gas at $a=V_a$
Volume of gas at $b=V_b$
$n=$ no. of moles $=2$
$R=8.314J\cdot K^{-1}\cdot mol^{-1}=$ Universal Gas constant.
Ideal gas law:
$PV_a=nRT_a=2RT_a=2R(200)=400R$
$PV_b=nRT_b=2R(300)=600R$
Work done in process $ab$ $=PV_b-PV_a=200R=(200\times 8.314)J=1662.8J$
Let $W=$ work done in process $ca$.
$W+1662.8=\Delta W=-800$
$\Longrightarrow W=(-800-1662.8)J=-2462.8J$
In the process $ca$, $2462.8J$ of work is done on the gas.
(Work done on the system is negative and work done by the system is positive)
