Question

Given that for a gas ${B_i}\, = $ isothermal bulk modulus, ${B_a}\, = $ adiabatic bulk modulus, pressure $ = \,p$ and $\gamma \, = $ the ratio of the specific heat at constant pressure to that constant volume. Which one of the following relations is correct?
(A) ${B_a}\, = \,p$
(B) ${B_i}\, = \,p$
(C) ${B_i}\, = \,\gamma {B_a}$
(D) None of the above

Answer 1

Hint
You can easily derive the relations between ${B_i},\,p$ , ${B_a},\,p$ and ${B_i},\,{B_a}$ using the relation of bulk modulus for isothermal and adiabatic processes and manipulating them to an extent. Eg: for isothermal processes, we can write
${B_i}\, = \, - V \times \dfrac{{dP}}{{dV}}$ , now manipulate it to find the desired relations.

Complete step by step answer
To solve this question, as told in the hint, we need to manipulate the basic conditions of the respective processes. But first, we need to know the basic conditions of isothermal and adiabatic processes.
For isothermal processes, we can define bulk modulus as:
${B_i}\, = \, - V \times \dfrac{{dP}}{{dV}}$
Similarly, the basic condition for adiabatic processes is:
$P{V^x}\, = $ constant
Another thing that we’ll need to us is the Ideal gas equation:
$PV\, = \,nRT$
Where, $R$ is the ideal gas constant,
$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$T$ is the temperature, and
$n$ is the number of moles of the gas.
First, let’s manipulate the equation of bulk modulus for isothermal processes:
We know that, ${B_i}\, = \, - V \times \dfrac{{dP}}{{dV}}$
And, $PV\, = \,nRT$
$P\, = \,\dfrac{{nRT}}{V}$
Now, we will differentiate the equation with respect to Volume, and we’ll get:
$\dfrac{{dP}}{{dV}}\, = \, - \dfrac{{nRT}}{{{V^2}}}$
Substituting this in the equation of Bulk modulus:
${B_i}\, = \, - V \times \,\dfrac{{ - nRT}}{{{V^2}}}$
${B_i}\, = \,\dfrac{{nRT}}{V}$
But $\dfrac{{nRT}}{V}\, = \,P$
So, we get:
${B_i}\, = \,P$
As we can see, option (B) is the correct answer.

Note
Although, it is not needed, but we are still finding a relation for adiabatic process, just to strengthen your concept:
$d\left( {P{V^x}} \right)\, = \,xP{V^{\left( {x - 1} \right)}}dV\, + \,{V^x}dP$
But we already know that for adiabatic process, $P{V^x}\, = $ constant
So, $d\left( {P{V^x}} \right)\, = \,0$
Hence,
$\Rightarrow xP{V^{\left( {x - 1} \right)}}dV\, + \,{V^x}dP\, = \,0$
After transposing, we reach at:
$\Rightarrow - \dfrac{{dV}}{V}\, = \,\dfrac{{dP}}{{xP}}$
After further transposing, we get:
$\therefore - V\dfrac{{dP}}{{dV}}\, = \,xP$ or ${B_a}\, = \,xP$.