Animated Solution for Physics - Thermodynamics: $C_p$ and $C_V$ are specific heats at constant pressure and constant volume, respectively. It is observed that $C_p-C_V=a$ for hydrogen gas $C_p-C_V=b$ for nitrogen gas. The correct relation between $a$ and $b$ is

Welcome, future engineers and scientists! Today, we are going to dissect a classic JEE problem that tests not just your mathematical prowess, but your ability to read carefully and avoid a very common trap.

The question asks us to find the relationship between the differences in specific heats for two gases: Hydrogen ($H_2$) and Nitrogen ($N_2$). At first glance, you might immediately think of Mayer's relation and confidently declare that the difference is always the universal gas constant, $R$. If you did that, you would conclude that $a=b$.

But wait! Let's take a breath and look closer. The question explicitly states that $C_p$ and $C_V$ are specific heats, not molar specific heats. This single word changes the entire landscape of the problem.

Let's clear up the terminology. Mayer's relation in its most famous form is written for one mole of an ideal gas:

Here, $C_{p,m}$ and $C_{V,m}$ are the molar specific heats (the heat required to raise the temperature of one mole of gas by one Kelvin).

However, the specific heat capacity (often denoted by lowercase $c_p$ and $c_v$, though our question uses capital letters to be tricky) is the heat required to raise the temperature of one unit mass (like one gram or one kilogram) of the gas by one Kelvin.

To convert from molar specific heat to specific heat per unit mass, we divide by the molar mass, $M$, of the gas:

Substituting these into Mayer's relation, we get our master equation for this problem:

Now that we have the correct tool, let's apply it to our two gases.

For Hydrogen ($H_2$): Hydrogen is a diatomic gas. Each molecule consists of two hydrogen atoms. Since the atomic mass of hydrogen is approximately $1\text{ g/mol}$, the molar mass of $H_2$ is $M_{H_2}=2\text{ g/mol}$.

Plugging this into our master equation, we get the value for $a$:

For Nitrogen ($N_2$): Nitrogen is also a diatomic gas. The atomic mass of nitrogen is $14\text{ g/mol}$, making the molar mass of $N_2$ equal to $M_{N_2}=28\text{ g/mol}$.

Plugging this into our master equation, we get the value for $b$:

We now have a system of two simple equations:

1. $a=\frac{R}{2}$ 2. $b=\frac{R}{28}$

Our goal is to find the relationship between $a$ and $b$. The easiest way to do this is to express $R$ in terms of $b$ from the second equation and substitute it into the first.

From the second equation, we multiply both sides by 28:

Now, substitute this expression for $R$ into the first equation:

Simplifying the fraction, we arrive at our final, elegant result:

And there we have it! By paying close attention to the physical definitions of the terms provided, we navigated around a dangerous trap and arrived safely at the correct answer. Always remember: in physics, the units and the precise definitions of the variables are just as important as the equations themselves.