Animated Solution for Physics - Thermodynamics: A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process?

Thermodynamics is often perceived as a dense forest of equations, but at its core, it is simply the accounting of energy. Imagine a bank account where heat is the income, work is the expense, and internal energy is your savings balance. In this problem, we are given a specific transaction: a diatomic gas expands at a constant pressure, doing $10\text{ J}$ of work. Our goal is to find out how much heat was deposited into the system to make this happen.

We are told the gas is expanding at a constant pressure. In the language of thermodynamics, this is an isobaric process. The work done by a gas expanding against a constant pressure $p$ is given by the simple relation:

However, we can also view this through the lens of the Ideal Gas Law, $pV=nRT$. Since the pressure $p$ is constant, any change in volume $\Delta V$ must be directly proportional to a change in temperature $\Delta T$. Therefore, we can rewrite the work done as:

We are given that $W=10\text{ J}$. This means we have a powerful piece of information: $nR\Delta T=10\text{ J}$. We don't need to know the number of moles $n$, the gas constant $R$, or the exact temperature change $\Delta T$ individually. Their combined product is all we need!

Now, let's talk about the heat absorbed. For an isobaric process, the heat exchanged is governed by the molar heat capacity at constant pressure, denoted as $C_p$. The formula is:

To proceed, we need the value of $C_p$. This is where the nature of the gas comes into play. The problem specifies a diatomic gas with rigid molecules. A rigid diatomic molecule (like a tiny dumbbell) can move in three dimensions (3 translational degrees of freedom) and rotate around two independent axes (2 rotational degrees of freedom). It does not vibrate because it is 'rigid'. This gives it a total of $f=5$ degrees of freedom.

The molar heat capacity at constant volume $C_v$ is given by $\frac{f}{2}R$, which means $C_v=\frac{5}{2}R$. Using Mayer's relation ($C_p=C_v+R$), we find:

Let's substitute our $C_p$ back into the heat equation:

By slightly rearranging the terms, we can group the variables we already know:

Remember that golden nugget of information we saved earlier? We know that $nR\Delta T=10\text{ J}$. Substituting this directly into our equation yields:

And there we have it! The gas absorbed $35\text{ J}$ of heat energy. Out of this $35\text{ J}$, it spent $10\text{ J}$ doing work on its surroundings, and the remaining $25\text{ J}$ went into increasing its own internal energy (its temperature). The beauty of thermodynamics lies in this perfect, unbreakable balance.