Question

Find the molar specific heat (in $\text{J mol}{}^{-1}\text{K}{}^{-1}$) for a process where $P=\frac{a}{T}$, for one mole of a monatomic gas, ${}^{\prime }{a}^{\prime }$ being constant.

• A. $\frac{5}{2}R$
• B. $\frac{7}{2}R$
• C. $R$
• D. $2R$

Answer 1

The correct option is B $\frac{7}{2}R$

We know that, from first law of thermodynamics.

$dQ=dU+dW$

Specific heat, $C=\frac{dQ}{dT}=\frac{dU}{dT}+\frac{dW}{dT}$

Since, $dU=C_{v}dT$ for 1 mole of gas,

$C=C_v+\frac{dW}{dT}$

$C=C_v+\frac{PdV}{dT}......\left(1\right)$

Also, $PV=RT[\because n=1]$

For the given process,

$V=\frac{RT}{P}=\frac{R{T}^2}{a}[\because P=\frac{a}{T}]$

$\Rightarrow \frac{dV}{dT}=\frac{2RT}{a}$

Using this in $\left(1\right)$

So, $C=C_v+P\left(\frac{2RT}{a}\right)$

$\Rightarrow C=C_v+2R[\because P=\frac{a}{T}]$

$\Rightarrow C=\frac{3}{2}R+2R$

[$C_v=\frac{3}{2}R$ for monoatomic gas]

$\Rightarrow C=\frac{7R}{2}$