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The ratio of the molar heat capacity of a diatomic gas at constant pressure to that at constant volume is?
$\begin{align}
  & \text{A}\text{. }\dfrac{7}{5} \\
 & \text{B}\text{. }\dfrac{3}{2} \\
 & \text{C}\text{. }\dfrac{3}{5} \\
 & \text{D}\text{. }\dfrac{5}{2} \\
\end{align}$

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Last updated date: 24th Jul 2024
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Answer
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Hint: Define molar heat capacity. Obtain the relation between the heat capacity at constant pressure and at constant volume. Find the heat capacity at constant pressure for a diatomic gas. Putting this value on the relation for the heat capacity at constant pressure and constant volume. Take the ratio of these two to get the required answer.

Complete step by step answer:
Heat capacity of a material can be defined as the amount of heat that should be supplied to a material to increase its temperature by a unit amount.
The specific heat capacity of a material or a substance can be defined as the heat required to increase the temperature of the unit mass of the substance by one degree or one unit is called the specific heat capacity of the substance.
The molar heat capacity of a material or a substance can be defined as the heat required to increase the temperature of one mol of the substance by one degree or one unit is called the molar heat capacity of the substance.
We have a diatomic gas.
Now, specific heat at constant volume of an ideal gas is, ${{C}_{v}}=\dfrac{5}{2}R$
So, the specific heat at constant pressure of the diatomic gas will be,
$\begin{align}
  & {{C}_{p}}={{C}_{V}}+R \\
 & {{C}_{p}}=\dfrac{5}{2}R+R \\
 & {{C}_{p}}=\dfrac{7}{2}R \\
\end{align}$
The ratio of the molar heat capacity of a diatomic gas at constant pressure to that at constant volume is,
$\dfrac{{{C}_{p}}}{{{C}_{v}}}=\dfrac{\dfrac{7}{2}R}{\dfrac{5}{2}R}=\dfrac{7}{5}$

The correct option is (A).

Note:
For different gases the specific heat capacity will be different. For a monatomic gas, the specific heat capacity at constant volume is $\dfrac{3}{2}R$. For a polyatomic gas, the specific heat capacity at constant volume is $3R$.