Question

An ideal monoatomic gas is at pressure $P_0$ and volume $V_0$. It is taken to final volume $2V_0$ and final pressure $\frac{P_0}{2}$ in a process which is straight line on $P-V$ diagram.

• A. Final temperature is greater than initial temperature
• B. Internal energy increases
• C. Work done by the gas is $+\frac{P_0{V}_0}{4}$
• D. Heat is absorbed by the gas

Answer 1

The correct option is D Heat is absorbed by the gas

Initial temperature,
$T_i=\frac{P_0{V}_0}{nR}$
and
Final temperature,
$T_f=\frac{\frac{P_0}{2}\times 2V_0}{nR}=\frac{P_0{V}_0}{nR}$
So, $T_i=T_f$
As, $T_i=T_f,\Delta U=0$
Workdone $=$ area under $P-V$ graph
$=\frac{1}{2}\times (P_0+P_0/2)\times (2V_0-V_0)=\frac{3P_0{V}_0}{4}$


As, $\Delta U=0$ and $W>0$
$\Rightarrow \Delta Q>0$
[ From first law of thermodynamics ]
Therefore, heat is absorbed in the process.