Question

An ideal gas $\left( {{{C}_{P}}}/{{{C}_{V}}=\gamma }\; \right)$ is taken through a process in which the pressure and the volume vary as $P=a{{V}^{b}}$. Find the value of b for which the specific heat capacity in the process is zero.
$\begin{align}
  & \text{A}\text{. }b=2\gamma \\
 & \text{B}\text{. }b=-\gamma \\
 & \text{C}\text{. }b=\dfrac{1}{\gamma } \\
 & \text{D}\text{. }b=-\dfrac{1}{\gamma } \\
\end{align}$

Answer 1

Hint: Relation of heat exchange, work done and internal energy is given by the first law of thermodynamics. Use the principle of conservation of energy, provided a condition that specific heat is zero. Again solve for internal energy using the ideal gas equation and relation ${{C}_{P}}-{{C}_{V}}=R$. Compare two equations will get an answer of b.

Formula used:
According to first law of thermodynamics
$\Delta Q=\Delta U+\Delta w$
Where,
$\Delta Q=$ Heat supplied to system by the surrounding
$\Delta w=$ Work done by system on surroundings
$\Delta U=$ Change in internal energy.

Complete step by step answer:
An ideal gas $\left( {{{C}_{P}}}/{{{C}_{V}}=\gamma }\; \right)$ is taken a process in which the pressure and volume vary as $P=a{{V}^{b}}$.
Given,
$\begin{align}
  & \dfrac{{{C}_{P}}}{{{C}_{V}}}=\gamma \\
 & \Rightarrow \dfrac{{{C}_{P}}}{{{C}_{V}}}-1=\gamma -1 \\
 & \Rightarrow \dfrac{{{C}_{P}}-{{C}_{V}}}{{{C}_{V}}}=\gamma -1 \\
 & \Rightarrow \dfrac{R}{\gamma -1}={{C}_{V}} \\
\end{align}$
i.e. $\dfrac{nR}{\gamma -1}={{C}_{V}}$ for n mole
It is given that the specific heat capacity is zero. That means heat (Q) must be zero. When $Q=0$ Then the process is called an adiabatic process. According to principle of conservation of energy
$\Delta Q=\Delta U+\Delta w$
Where,
 $\Delta Q=$ Heat supplied to system by the surrounding
$\Delta w=$ Work done by system on surroundings
$\Delta U=$ Change in internal energy.
$\therefore \Delta w=-\Delta U$
Thus, $\Delta w=-\Delta U=\int{-PdV}$
Put value of $P=a{{V}^{b}}$
$w=\int{a{{V}^{b}}dV=-\dfrac{a}{b+1}\left( V_{2}^{b+1}-V_{1}^{b+1} \right)...............\left( 1 \right)}$
Since, $D=a{{V}^{b}},$ So $P=a{{V}^{b}}$ again in equation (1)
Therefore, $w=-\dfrac{{{P}_{2}}{{V}_{2}}-{{P}_{1}}{{V}_{1}}}{b+1}...........(4)$
According to ideal gas equation, $Pv=nRT$ using equation (Q)
$\Delta U=\dfrac{nR}{\gamma -1}\left( {{T}_{2}}-{{T}_{1}} \right)....................\left( 2 \right)\left( \because {{C}_{V}}=\dfrac{\Delta U}{\Delta T} \right)$
If an ideal gas undergoes a change in its state adiabatically from $\left( {{P}_{1}},{{V}_{1}} \right)$ to $\left( {{P}_{2}},{{V}_{2}} \right)$ then equation (@) is written as –
$\Delta U=\dfrac{{{P}_{2}}{{U}_{2}}-{{P}_{1}}{{V}_{1}}}{\gamma -1}...........\left( 3 \right)$
Compare equation (3) & (4), we get –
$b+1=1-\gamma $
Thus $b=-\gamma $.
Therefore, the correct option is (B).

Note:
This question can be solved in different ways. We have provided that the specific heat capacity is zero i.e. it must be the adiabatic process.
Adiabatic process is nothing but the system is insulated from the surroundings and heat absorbed is related must be equal to zero.
For adiabatic process of an ideal gas, expression of pressure and volume is gives as –
$\begin{align}
  & PV\gamma =\text{Constant} \\
 & PV\gamma =a \\
 & P=a{{V}^{-\gamma }}.........\left( i \right) \\
\end{align}$
We have provided $P=a{{V}^{b}}...........\left( ii \right)$
Compare (i) & (ii), we get –
$b=-\gamma $