Animated Solution for Physics - Thermodynamics: 70 cal of heat are required to raise the temperature of 2 moles of an ideal diatomic gas at constant pressure from 30°C to 35°C. The amount of heat required (in calorie) to raise the temperature of the same gas through the same range (30°C to 35°C) at constant volume is

The problem presents us with a classic thermodynamic scenario: heating a gas under two different constraints—constant pressure and constant volume. We are given that $70\text{ cal}$ of heat is required to raise the temperature of $2\text{ moles}$ of an ideal diatomic gas by $5^{\circ }\text{C}$ (from $30^{\circ }\text{C}$ to $35^{\circ }\text{C}$) at constant pressure. Our goal is to find the heat required to achieve the exact same temperature rise for the same gas, but this time keeping the volume constant.

Imagine two identical cylinders filled with this diatomic gas. In the first cylinder, the piston is free to move, maintaining a constant pressure as the gas expands. In the second cylinder, the piston is locked in place, keeping the volume strictly constant. We know the heat supplied to the first cylinder, and we need to find the heat for the second.

Let's write down the fundamental equations for heat transfer in these two processes. The heat supplied at constant pressure is given by:

Similarly, the heat supplied at constant volume is:

Here, $n$ is the number of moles, $C_p$ and $C_V$ are the molar heat capacities at constant pressure and constant volume respectively, and $\Delta T$ is the change in temperature.

We could calculate $C_p$ directly since we know $Q_p$, $n$, and $\Delta T$. However, there is a much more elegant and faster approach. Let's take the ratio of the two heat equations:

Notice how beautifully the terms $n$ and $\Delta T$ cancel out! This tells us that for the same gas undergoing the same temperature change, the ratio of the heats is simply the ratio of their molar heat capacities:

We know that the ratio of specific heats, $\gamma$, is defined as $\gamma =\frac{C_p}{C_V}$. Therefore, our equation simplifies to:

The problem explicitly states that we are dealing with an ideal diatomic gas. For a diatomic gas at normal temperatures, the molecules have 3 translational and 2 rotational degrees of freedom, giving a total of $f=5$ degrees of freedom.

Using the equipartition theorem, the molar heat capacity at constant volume is:

And the molar heat capacity at constant pressure is:

The adiabatic exponent $\gamma$ is their ratio:

Now, we simply substitute the known values into our elegant ratio equation. We know $Q_p=70\text{ cal}$ and $\gamma =1.4$.

And there we have it! The heat required at constant volume is $50\text{ cal}$.

This result makes perfect physical sense. At constant volume, all the supplied heat goes entirely into increasing the internal energy of the gas. However, at constant pressure, the gas expands and does work on its surroundings. Therefore, extra heat must be supplied to account for this work, which is why $Q_p$ ($70\text{ cal}$) is greater than $Q_V$ ($50\text{ cal}$).