Question

The diagram shows the adiabatic curve for $n$ moles of an ideal gas. The bulk modulus for the gas corresponding to the point $P$ will be

• A. $\frac{5 \, \textit{nRT}_{0}}{3 \textit{V}_{0}}$
• B. $\textit{nR}\left(2 + \frac{\left(\textit{T}\right)_{0}}{\left(\textit{V}\right)_{0}}\right)$
• C. $\textit{nR}\left(1 + \frac{\left(\textit{T}\right)_{0}}{\left(\textit{V}\right)_{0}}\right)$
• D. $\frac{2 \, \textit{nRT}_{0}}{\textit{V}_{0}}$

Answer 1

Answer: (D)

For adiabatic process :
Bulk modulus : $\text{B} = \gamma \text{ P}$
For point P : $\text{P} = \frac{\text{nRT}}{\text{V}} = \frac{\text{nR} 3 \text{T}_{0}}{3 \text{V}_{0}} = \frac{\text{nRT}_{0}}{\text{V}_{0}}$
$⇒ \, \, \, \text{B} = \frac{\gamma \text{nRT}_{0}}{\text{V}_{0}}$ ...(i)
Now, $\text{TV}^{\gamma - 1} = \text{constant}$
$⇒ \, \, \, \left(\gamma - 1\right) \text{TdV} + \text{VdT} = 0$
$⇒ \, \, \, \frac{\text{dV}}{\text{dT}} = \frac{- \text{V}}{\left(\gamma - 1\right) \text{T}}$
For point P $ \, \, \rightarrow \, $ $\frac{- 3 \left(\text{V}\right)_{0}}{3 \left(\text{T}\right)_{0}} = \frac{- \left(3 \left(\text{V}\right)_{0}\right)}{\left(\gamma - 1\right) \left(3 \left(\text{T}\right)_{0}\right)}$
$⇒ \, \, \, \gamma = 2$
So, from Eq. (i), $\text{B} = \frac{2 \text{nRT}_{0}}{\text{V}_{0}}$