Animated Solution for Physics - Thermodynamics: 5.6 L of helium gas at STP is adiabatically compressed to 0.7 L. Taking the initial temperature to be $T_1$, the work done in the process is

Imagine you are standing next to a perfectly insulated cylinder filled with Helium gas. You push the piston down, compressing the gas from a comfortable $5.6\text{ L}$ to a tight $0.7\text{ L}$. Because the cylinder is insulated, no heat can escape. This is a classic adiabatic compression.

Our mission is to find out exactly how much work you had to do to compress this gas. Let's break this down step-by-step.

Finding the Moles

Before we can use any fancy thermodynamic equations, we need to know how much gas is actually in the cylinder. The problem gives us a massive clue: the gas is initially at STP (Standard Temperature and Pressure).

At STP, we know that exactly $1\text{ mole}$ of any ideal gas occupies a volume of $22.4\text{ L}$. This is a universal constant you should always keep in your back pocket!

Since our initial volume $V_1$ is $5.6\text{ L}$, we can easily find the number of moles $n$:

So, we are dealing with exactly a quarter mole of Helium.

Temperature and Volume

In an adiabatic process, the temperature and volume of an ideal gas are intimately connected by the relation:

This means that $T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$.

To use this, we need $\gamma$, the adiabatic index. Helium is a noble gas, meaning it is monoatomic. For monoatomic gases, the degrees of freedom $f=3$, which gives us $\gamma =1+\frac{2}{f}=1+\frac{2}{3}=\frac{5}{3}$.

Now, let's rearrange our equation to solve for the final temperature $T_2$:

Substituting our known values:

Don't let the fractional powers scare you! The math simplifies beautifully:

Since $8$ is just $2^3$, we have:

The temperature has quadrupled! This makes perfect physical sense: when you compress a gas adiabatically, you are doing work on it, which increases its internal energy and thus its temperature.

Calculating the Work Done

We now have all the pieces of the puzzle. The work done $W$ by an ideal gas in an adiabatic process is given by the master equation:

Let's plug in our hard-earned values: $n=\frac{1}{4}$, $T_2=4T_1$, and $\gamma =\frac{5}{3}$.

The negative sign is crucial here. It tells us that work was done on the gas (compression), rather than by the gas (expansion).

Since the question asks for the work done in the process and provides positive options, it is asking for the magnitude of the work done on the gas. Therefore, the magnitude is $\frac{9}{8}R{T}_1$, which perfectly matches option (a).