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. $\frac{5{\text{nRT}}_0}{3{\text{V}}_0}$
• B. $\text{nR}\left(2+\frac{{\left(\text{T}\right)}_0}{{\left(\text{V}\right)}_0}\right)$
• C. $\text{nR}\left(1+\frac{{\left(\text{T}\right)}_0}{{\left(\text{V}\right)}_0}\right)$
• D. $\frac{2{\text{nRT}}_0}{{\text{V}}_0}$

Answer 1

Answer: (D)

For adiabatic process :
Bulk modulus : $\text{B}=\gamma \text{ P}$
For point P : $\text{P}=\frac{\text{nRT}}{\text{V}}=\frac{\text{nR}3{\text{T}}_0}{3{\text{V}}_0}=\frac{{\text{nRT}}_0}{{\text{V}}_0}$
$\Rightarrow \text{B}=\frac{\gamma {\text{nRT}}_0}{{\text{V}}_0}$ ...(i)
Now, ${\text{TV}}^{\gamma -1}=\text{constant}$
$\Rightarrow \left(\gamma -1\right)\text{TdV}+\text{VdT}=0$
$\Rightarrow \frac{\text{dV}}{\text{dT}}=\frac{-\text{V}}{\left(\gamma -1\right)\text{T}}$
For point P $\to$ $\frac{-3{\left(\text{V}\right)}_0}{3{\left(\text{T}\right)}_0}=\frac{-\left(3{\left(\text{V}\right)}_0\right)}{\left(\gamma -1\right)\left(3{\left(\text{T}\right)}_0\right)}$
$\Rightarrow \gamma =2$
So, from Eq. (i), $\text{B}=\frac{2{\text{nRT}}_0}{{\text{V}}_0}$