Question

If isothermal bulk modulus is $x$ times adiabatic bulk modulus, then value of $x$ is equal to

• A. $\gamma$
• B. $\frac{1}{\gamma }$
• C. ${\gamma }^2$
• D. $\frac{1}{{\gamma }^2}$

Answer 1

The correct option is B $\frac{1}{\gamma }$

For an isothermal process,
Equation of state is $PV=k$
Differentiating on both sides
$PdV+VdP=0$
$\therefore$ Bulk modulus at constant temperature,
$k_T=\frac{-dP}{\left(\frac{dV}{V}\right)}=P$
For an adiabatic process,
Equation of state is $P{V}^{\gamma }=k$
Differentiating on both sides,
$P\left(\gamma V^{\gamma -1}dV\right)+V^{\gamma }dP=0$
$\therefore$ Adiabatic Bulk modulus $\left(k_s\right)=\frac{-dP}{\left(\frac{dV}{V}\right)}=\gamma P$
Given that $k_T=x{k}_s$
$P=x\left(\gamma P\right)$
$\Rightarrow x=\frac{1}{\gamma }$