Question

The molar specific heats of an ideal gas at constant pressure and volume are denoted by ${{C}_{P}}$ and ${{C}_{V}}$ respectively. If $\gamma =\dfrac{{{C}_{P}}}{{{C}_{V}}}$ and $R$ is the universal gas constant, then ${{C}_{V}}$ is equal to:
$\begin{aligned} & \text{A}\text{. }\dfrac{\left( \gamma -1 \right)}{R} \\\ & \text{B}\text{. }\gamma R \\\ & \text{C}\text{. }\dfrac{1+\gamma }{1-\gamma } \\\ & \text{D}\text{. }\dfrac{R}{\left( \gamma -1 \right)} \\\ \end{aligned}$

Answer 1

Specific heat capacity of a gas is the amount of heat required to raise the temperature of one gram gas by unit degree but per mole of the gas is called molar heat capacity or simply heat capacity. Usually, heat capacity equation expressed at constant pressure $\left( {{C}_{P}} \right)$ and constant volume $\left( {{C}_{V}} \right)$. We will find the relation between $\left( {{C}_{P}} \right)$, $\left( {{C}_{V}} \right)$and $\gamma$ with the help of universal gas constant $R$.

Complete step by step answer:
The specific heat of a gas is the amount of energy required to raise the temperature of one mole of gas by one kelvin. The reason gasses have two specific heats defined because they are not stable; they change more than liquids and solids. Therefore, when the volume of the gas is held constant we define the heat capacity at constant volume $\left( {{C}_{V}} \right)$ and when the pressure of the gas is held constant we define the heat capacity at constant pressure$\left( {{C}_{P}} \right)$.
The formula $q=nC\Delta t$ represents the heat $q$ required to bring about a $\Delta t$ difference in temperature of one mole of any matter. The constant $C$ here represents the molar heat capacity of the body.
The molar heat capacity of a matter is defined as the amount of heat energy required to change, increase or decrease, the temperature of one mole of that substance by one unit. This value depends on the nature, size, and composition of the system.
From the equation $q=nC\Delta t$, we can say that,
At constant pressure $P$, we have
${{q}_{P}}=n{{C}_{P}}\Delta t$
This value is equal to the change in enthalpy of the gas, that is,
${{q}_{P}}=n{{C}_{P}}\Delta t=\Delta H$
Similarly, at constant volume $V$, we have
${{q}_{V}}=n{{C}_{V}}\Delta t$
This value is equal to the change in internal energy of the gas, that is,
${{q}_{V}}=n{{C}_{V}}\Delta t=\Delta U$
As we know,
For one mole of an ideal gas,
$\begin{aligned} & \Delta H=\Delta U+\Delta \left( PV \right) \\\ & \Delta H=\Delta U+\Delta \left( Rt \right) \\\ & \Delta H=\Delta U+R\Delta t \\\ \end{aligned}$
Therefore,
$\begin{aligned} & {{C}_{P}}\Delta t={{C}_{V}}\Delta t+R\Delta t \\\ & {{C}_{P}}={{C}_{V}}+R \\\ & {{C}_{P}}-{{C}_{V}}=R \\\ \end{aligned}$
Now,
Let’s say equation 1, $\dfrac{{{C}_{P}}}{{{C}_{V}}}=\gamma$
Let’s say equation 2, ${{C}_{P}}-{{C}_{V}}=R$
From equation 1, we have,
${{C}_{P}}=\gamma {{C}_{V}}$
Put above value of ${{C}_{P}}$ in equation 2,
$\begin{aligned} & \gamma {{C}_{V}}-{{C}_{V}}=R \\\ & {{C}_{V}}\left( \gamma -1 \right)=R \\\ & {{C}_{V}}=\dfrac{R}{\left( \gamma -1 \right)} \\\ \end{aligned}$
The value of ${{C}_{V}}$ is $\dfrac{R}{\left( \gamma -1 \right)}$

So, the correct answer is “Option D”.

Note:
Students should remember that ${{C}_{V}}$ is the amount of heat energy that a substance or system absorbs or releases, per unit mass, with the change in temperature where volume change of the system does not occur while ${{C}_{P}}$ is the amount of heat energy that a substance or system absorbs or release, per unit mass, with the change in temperature where pressure change of the system does not occur.