Question

Figure shown below is a $PT$ graph of a monoatomic gas which undergoes a cyclic process. If work done by gas in process $BC$ is $x$, the heat absorbed in process $CA$ is

• A. $xln2$
• B. $-xln2$
• C. $\frac{x}{ln2}$
• D. $\frac{-x}{ln2}$

Answer 1

The correct option is A $xln2$

From the graph,
$BC$ is isobaric.
So, Workdone $W=nR(T_0-2T_0)$
$\Rightarrow W=x=-nR{T}_0$

Also, $CA$ is isothermal.
So, Heat $Q_{ca}=nR{T}_{0}ln\left(\frac{P_0}{2P_0}\right)$
$Q_{ca}=(-x)(-ln(2\left)\right)$
$Q_{ca}=xln2$