Specific heat capacity of a gas: Consider a mole of gas is heated in the cylinder with frictionless piston, then the change in internal energy of the gas is given by the first law of thermodynamics
(10.6.1)
\begin{equation*}
\,dQ=\,dU+\,dW
\end{equation*}
where
\(\,dW \) is the infinitesimal work done by the gas in moving a piston from initial position
\(i\) to the final position
\(f\text{,}\) and is given by
\begin{equation*}
\,dW = F\,dx= pA\,dx= p\,dV
\end{equation*}
where
\(A\) is area of cross-section of the piston,
\(F\) is force applied by the gas molecules on the surface of the piston.
\begin{equation}
\,dQ=\,dU+ p\,dV \tag{10.6.2}
\end{equation}
If the piston is fixed from moving, then at constant volume,
\(\,dV = 0\text{,}\) and
\begin{equation*}
\,dW = p\,dV = 0
\end{equation*}
\begin{equation}
\therefore \quad \,dQ=\,dU \tag{10.6.3}
\end{equation}
But, for n mole of a gas
\begin{equation*}
\,dQ=n C\,dT
\end{equation*}
where C is molar specific heat capacity of the gas.
\begin{equation*}
\,dQ=mc\,dT
\end{equation*}
where
\(m\) is mass of the gas, but it is convenient to take mole instead of mass as an amount of gas. At constant volume,
\begin{equation}
\,dQ=n C_{v}\,dT \tag{10.6.4}
\end{equation}
From eqns.
(10.6.3) and
(10.6.4), we have -
\begin{equation}
\frac{\,dU}{\,dT}=nC_{v} \tag{10.6.5}
\end{equation}
At constant pressure,
\begin{equation}
\,dQ=n C_{p}\,dT \tag{10.6.6}
\end{equation}
\begin{equation}
\therefore\quad \frac{\,dQ}{\,dT}=nC_{p} \tag{10.6.7}
\end{equation}
Now, differentiating eqn.
(10.6.2) with respect to
\(T\text{,}\) we get -
\begin{equation}
\frac{\,dQ}{\,dT} = \frac{\,dU}{\,dT} +p\frac{\,dV}{\,dT} \tag{10.6.8}
\end{equation}
But from euation of state of a gas
(10.5.8),
\begin{equation*}
pV = nRT
\end{equation*}
Differentiating this equation with respect to
\(T\) at constant
\(P\text{,}\) we get -
\begin{equation}
\text{or,}\quad p\frac{\,dV}{\,dT} = nR \tag{10.6.9}
\end{equation}
substituting the values in eqn.
(10.6.8) from eqns.
(10.6.5) and
(10.6.7), we get -
\begin{equation*}
nC_{p}=nC_{v}+nR
\end{equation*}
\begin{equation}
\therefore\quad C_{p}-C_{v} = R \tag{10.6.10}
\end{equation}
This is the relation between molar specific heat capacity of a gas at constant pressure
\(C_{p}\) and molar specific heat capacity of a gas at constant volume
\(C_{v}\text{.}\) From the kinetic theory of gas, the internal energy of a monoatomic gas [
(10.5.11)] is given by
\begin{equation*}
U=\frac{3}{2}nRT
\end{equation*}
\begin{equation}
\therefore\quad \frac{\,dU}{\,dT} = \frac{3}{2}nR \tag{10.6.11}
\end{equation}
Hence, from eqns.
(10.6.8),
(10.6.9), and
(10.6.11), we have -
\begin{equation*}
\frac{\,dQ}{\,dT} =\frac{3}{2}nR + nR =\frac{5}{2}nR
\end{equation*}
Now, the ratio of the specific heat at constant pressure to the specific heat at constant volume of a monoatomic gas is given by
\begin{equation*}
\gamma = \frac{C_{p}}{C_{v}} = \frac{\frac{5}{2}nR}{\frac{3}{2}nR} = \frac{5}{3}
\end{equation*}
The idea of degrees of freedom can be extended to diatomic and polytomic gases. It can be shown that the value of
\(\gamma\) can be expressed as
\begin{equation*}
\gamma=\frac{f+2}{f}
\end{equation*}
where
\(f\) is the number of degrees of freedom per molecule. For a monatomic gas, we find that
\begin{equation*}
\gamma = \frac{3+2}{3} = \frac{5}{3}
\end{equation*}
since
\(f=3\) for a monoatomic gas.
\begin{equation*}
\gamma = \frac{5+2}{3} = \frac{7}{3}
\end{equation*}
for a diatomic gas as
\(f=5\) .
