Question

If $'{{C}_{p}}'$ and $'{{C}_{v}}'$ are molar specific heats of an ideal gas at constant pressure and volume respectively, $'\gamma '$ is the ratio of the two specific heats and $'R'$ is the universal gas constant, then $'{{C}_{p}}'$ is equal to?
(A). $\dfrac{R\gamma }{\gamma -1}$
(B). $\gamma R$
(C). $\dfrac{1+\gamma }{1-\gamma }$
(D). $\dfrac{R}{\gamma -1}$

Answer 1

Hint: This problem can be solved by using the relation between the specific heats at constant pressure and volume and the universal gas constant. The answer can be found out by plugging in the given value of the ratio of the two specific heats.

Formula used:
${{C}_{p}}-{{C}_{v}}=R$
where ${{C}_{p}}$ and ${{C}_{v}}$ are the molar specific heat capacities at constant pressure and volume of a gas respectively and $R$ is the universal gas constant equal to $8.314J.mo{{l}^{-1}}{{K}^{-1}}$.

Complete step by step answer:
As explained in the hint, we will use the relation between the molar specific heats at constant pressure and volume and the universal gas constant to find out the required value in terms of the ratio of the specific heats given in the question.
Hence, let us proceed.
Now,
${{C}_{p}}-{{C}_{v}}=R$ --(1)
where ${{C}_{p}}$ and ${{C}_{v}}$ are the molar specific heat capacities at constant pressure and volume of a gas respectively and $R$ is the universal gas constant equal to $8.314J.mo{{l}^{-1}}{{K}^{-1}}$.
Also, it is given in the question that the ratio of the two specific heats is $\gamma $.
$\therefore \dfrac{{{C}_{p}}}{{{C}_{v}}}=\gamma $
$\therefore {{C}_{v}}=\dfrac{{{C}_{p}}}{\gamma }$ --(2)
Putting (2) in (1), we get,
${{C}_{p}}-\dfrac{{{C}_{p}}}{\gamma }=R$
$\therefore {{C}_{p}}\left( 1-\dfrac{1}{\gamma } \right)=R$
$\therefore {{C}_{p}}\left( \dfrac{\gamma -1}{\gamma } \right)=R$
$\therefore {{C}_{p}}=R\left( \dfrac{\gamma }{\gamma -1} \right)$
$\therefore {{C}_{p}}=\dfrac{R\gamma }{\gamma -1}$
Hence, the required value of ${{C}_{p}}$ is $\dfrac{R\gamma }{\gamma -1}$.
Therefore, the correct option is A) $\dfrac{R\gamma }{\gamma -1}$.

Note: Students must properly know equation (1), since it is one of the most fundamental and important equations in thermodynamics regarding gases. Students also sometimes get confused whether the ratio is $\dfrac{{{C}_{p}}}{{{C}_{v}}}=\gamma $ or $\dfrac{{{C}_{v}}}{{{C}_{p}}}=\gamma $. An easy way to remember this is to keep in mind that the heat taken in a constant pressure process is always more than the heat taken in a constant volume process because no mechanical work is done by the gas in a constant volume process while in a constant pressure process, mechanical work is also done by the gas as there is a change in its volume. Hence, more heat is required for the process and hence, the molar specific heat capacity will also be larger. That being said, they must remember that $\gamma >1$ always. Hence, keeping these two things in mind, students will not get confused about the ratio.