Animated Solution for Physics - Thermodynamics: A cylinder with fixed capacity of 67.2 L contains helium gas at STP. The amount of heat needed to raise the temperature of the gas by 20°C is [Take, $R=8.31{\text{ J mol}}^{-1}{\text{K}}^{-1}$]

Imagine a rigid, unyielding steel cylinder. Inside this cylinder, we have a fixed volume of $67.2\text{ L}$ filled with Helium gas. The gas is initially at Standard Temperature and Pressure (STP). Our mission is to find out exactly how much heat energy we need to pump into this cylinder to raise the temperature of the Helium gas by $20^{\circ }\text{C}$.

Before we dive into the thermodynamics, we need to know exactly how much gas we are dealing with. The problem gives us the volume at STP.

One of the most beautiful shortcuts in chemistry and physics is the behavior of ideal gases at STP. At Standard Temperature and Pressure, exactly $1\text{ mole}$ of any ideal gas occupies a volume of $22.4\text{ L}$.

Since our cylinder holds $67.2\text{ L}$, we can easily find the number of moles ($n$) by dividing our total volume by the molar volume:

So, we are heating exactly $3\text{ moles}$ of Helium.

The phrase "fixed capacity" is the linchpin of this problem. It tells us that the walls of the cylinder cannot expand or contract. In thermodynamic terms, this is an isochoric process (constant volume).

Because the volume doesn't change, the gas does absolutely zero work on its surroundings ($W=0$). According to the First Law of Thermodynamics, all the heat ($\Delta Q$) we add goes directly into increasing the internal energy of the gas. The formula for heat added at constant volume is:

To use our heat formula, we need $C_V$, the molar heat capacity at constant volume. This value depends entirely on the atomic structure of the gas.

Helium is a noble gas, meaning it exists as single, independent atoms. It is monoatomic. A monoatomic gas only has $3$ translational degrees of freedom. Therefore, its molar heat capacity at constant volume is:

Now, we have all the pieces of the puzzle. Let's substitute our known values into the heat equation. We have $n=3$, $C_V=\frac{3}{2}R$, and $\Delta T=20^{\circ }\text{C}$.

Let's simplify the math before plugging in the messy decimal value for $R$. The $2$ in the denominator cancels with the $20$ to leave $10$.

Finally, we substitute the given value for the universal gas constant, $R=8.31{\text{ J mol}}^{-1}{\text{K}}^{-1}$:

Looking at our options, $747.9\text{ J}$ rounds perfectly to $748\text{ J}$. The correct option is (b).