Question

$V-T$ diagram for $n$ moles of monoatomic gas is given below:

Choose the correct statement(s).

• A. $\left|\frac{\Delta Q_{1-2}}{\Delta Q_{3-4}}\right|=\frac{1}{2}$
• B. $\left|\frac{\Delta Q_{1-2}}{\Delta Q_{2-3}}\right|=\frac{5}{3}$
• C. Work done in cyclic process is $\Delta W=\frac{nR{T}_0}{2}$
• D. There are only adiabatic and isochoric processes

Answer 1

Answer: (B), (C)

Work done in cyclic process is $\Delta W=\frac{nR{T}_0}{2}$

Process $1-2$:
If this line is extended then it will pass through origin which means $V\propto T$ so, $\frac{V}{T}=const.$
As $PV=nRT\Rightarrow \frac{V}{T}=\frac{nR}{P}$
In above expression if $P$ is constant then $\frac{V}{T}=\frac{nR}{P}=const.$
On comparision of both equations, it becomes clear that this line signifies constant pressure process.
So this is the isobaric expansion process.
Process 3-4:
Also constant pressure process but this is isobaric compression process.
Process 2-3:
It is a straight line parallel to the temperature axis. So this is a constant volume process.
Process 4-1:
It is a straight line parallel to the temperature axis and during the whole process, the volume is the same. So this process is also a constant volume process.

$\Delta Q_{1-2}=n{C}_p\Delta T$------(1)
$\Delta Q_{3-4}=n{C}_p\Delta T$--------(2)
From equation 1 and 2
$\left|\frac{\Delta Q_{1-2}}{\Delta Q_{3-4}}\right|=\left|\frac{n{C}_p{T}_0}{n{C}_p\frac{T_0}{2}}\right|=2$
Option A is incorrect

From equation 1 and 3
As, $\Delta Q_{2-3}=n{C}_v\Delta T$-----(3)
$\left|\frac{\Delta Q_{1-2}}{\Delta Q_{2-3}}\right|=\left|\frac{n{C}_p\Delta T}{n{C}_v\Delta T}\right|=\frac{C_p}{C_v}=\gamma =\frac{5}{3}$
Option B is correct

Work done during entire cycle will be,
$W_{1-2}+W_{2-3}+W_{3-4}+W_{4-1}$
Work done in $4-1$ and $2-3$ is zero as constant volume process
$W_{1-2}=p\Delta V=nR\Delta T=nR(2T_0-T_0)=nR{T}_0$
$W_{3-4}=p\Delta V=nR\Delta T=nR(\frac{T_0}{2}-T_0)$
$W_{3-4}=-nR\frac{T_0}{2}$
$W_{net}=W_{1-2}+W_{3-4}=nR\frac{T_0}{2}$
So , option C is correct
None of the process is adiabatic so option D is incorrect