Animated Solution for Physics - Thermodynamics: Comprehension Passage

The study of thermodynamics is essentially the study of how energy moves and transforms. In this beautiful problem, we are going to trace the journey of a helium gas sample as it undergoes two distinct thermodynamic processes. We will uncover the temperatures at various states, explore the profound concept of state functions, and finally, calculate the heat involved in each path. Let's dive in!

We are given a $2\text{ kg}$ sample of monoatomic helium gas. To apply the laws of thermodynamics, we first need to know how many particles we are dealing with, which means calculating the number of moles.

The mass of the gas is $m=2\text{ kg}=2000\text{ g}$. The molecular mass of helium is $M=4\text{ g/mol}$.

Now we have our system clearly defined: $500\text{ moles}$ of an ideal monoatomic gas.

Ideal Gas Law

To find the temperature at each state, we rely on the cornerstone of gas laws: the Ideal Gas Equation, $pV=nRT$. By rearranging this, we can solve for temperature: $T=\frac{pV}{nR}$.

Let's evaluate this for each state using the values from the $p-V$ graph.

For state $A$:

For state $B$:

For state $C$:

For state $D$:

Notice how the temperature scales directly with the product of pressure and volume!

Part (b) of the question asks a profound conceptual question: Can we tell which sample took path $ABC$ and which took path $ADC$ just by looking at them in their final state $C$?

The answer is a resounding No.

Why? Because the final state of both samples is exactly the same. Properties like temperature, pressure, volume, and internal energy are state functions. They depend exclusively on the current state of the system and have absolutely no memory of the path taken to get there. A gas at state $C$ is identical regardless of its history.

Heat, Work, and Energy

Now, let's calculate the heat involved in each process. The First Law of Thermodynamics states that the heat added to a system ($\Delta Q$) is the sum of the change in its internal energy ($\Delta U$) and the work done by the gas ($W$).

First, let's find the change in internal energy. Since internal energy is a state function, $\Delta U$ will be identical for both paths because they both start at $A$ and end at $C$. For a monoatomic gas, the molar heat capacity at constant volume is $C_V=\frac{3}{2}R$.

Now, let's calculate the work done, which is the area under the $p-V$ curve.

For path $ABC$: The process $A\to B$ is isochoric (constant volume), so no work is done. The process $B\to C$ is isobaric (constant pressure).

For path $ADC$: The process $A\to D$ is isobaric, and $D\to C$ is isochoric.

This beautifully illustrates that while internal energy is path-independent, heat and work are path-dependent. Different paths require different amounts of heat to achieve the exact same final state!