Question

A mono-atomic ideal gas undergoes a process in which the ratio of P to V at any instant is constant and is equal to $1$. What is the molar heat capacity of the gas?

• A. $2R$
• B. $\frac{3R}{2}$
• C. $\frac{5R}{2}$
• D. $0$

Answer 1

Answer: (A)

$2R$

Given that $P$ is equal to $V$.

For one mole of an ideal gas, $PV=RT$

$PdV+VdP=RdT$ or $2PdV=RdT$ (as $P=V$)

or, $PdV=\frac{RdT}{2}$ $...\left(1\right)$

Now, from first law of thermodynamics,

$dq=dU+PdV$

$dq=C_{v}dT+PdV$

$dq=C_{v}dT+\frac{RdT}{2}$ $($ from equation $1)$

For ideal monoatomic gas, $C_v=\frac{3R}{2}$

Integrating, we get $\frac{dq}{dT}=\frac{3R}{2}+\frac{R}{2}=\frac{4R}{2}=2R$

Hence, the correct option is $\text{A}$