Animated Solution for Physics - Thermodynamics: Two moles of helium gas undergo a cyclic process as shown in figure. Assuming the gas to be ideal, calculate the following quantities in this process. (a) The net change in the heat energy. (b) The net work done. (c) The net change in internal energy.

The world of thermodynamics is filled with fascinating cycles, and the problem we are tackling today is a perfect example of how elegant these cycles can be. We are presented with a $p-T$ (pressure-temperature) diagram for two moles of ideal helium gas undergoing a cyclic process $A\to B\to C\to D\to A$. Our goal is to find the net heat energy exchanged, the net work done, and the net change in internal energy.

I know this might look intimidating at first glance, but let's take a breath and break it down step-by-step. By analyzing each path individually, we will uncover the beautiful symmetry hidden within this cycle.

Let's start by looking at the horizontal lines on our graph: paths $A\to B$ and $C\to D$. Since the pressure remains constant along these lines, they represent isobaric processes.

In process $A\to B$, the gas is heated from $300\text{ K}$ to $400\text{ K}$ at a constant pressure of $2\text{ atm}$. The heat absorbed is given by $Q_{AB}=n{C}_p\Delta T_{AB}$.

Conversely, in process $C\to D$, the gas is cooled from $400\text{ K}$ back to $300\text{ K}$ at a constant pressure of $1\text{ atm}$. The heat released is $Q_{CD}=n{C}_p\Delta T_{CD}$.

Notice something interesting? The temperature change for $A\to B$ is $+100\text{ K}$, while for $C\to D$ it is $-100\text{ K}$. Because the number of moles $n$ and the molar heat capacity $C_p$ are the same for both processes, the heat absorbed during $A\to B$ perfectly cancels out the heat released during $C\to D$.

This is a massive simplification! We don't even need to calculate their exact values.

Now, let's focus on the vertical lines: paths $B\to C$ and $D\to A$. Here, the temperature is constant, making them isothermal processes.

For an ideal gas, the internal energy depends only on temperature. Since the temperature doesn't change during an isothermal process, the change in internal energy is zero ($\Delta U=0$). According to the First Law of Thermodynamics, this means all the heat added goes entirely into doing work ($Q=W$).

For process $B\to C$, the temperature is a constant $400\text{ K}$, and the pressure drops from $2\text{ atm}$ to $1\text{ atm}$. We can calculate the heat exchanged using the formula for isothermal work in terms of pressure:

Substituting our values ($n=2$, $R=8.31\text{ J/mol}\cdot \text{K}$, $T_B=400\text{ K}$):

Similarly, for process $D\to A$, the temperature is a constant $300\text{ K}$, and the pressure increases from $1\text{ atm}$ to $2\text{ atm}$.

The negative sign indicates that work is done on the gas, and heat is released to the surroundings.

We are now ready to find the net heat energy exchanged in the entire cycle. We simply sum the heat from all four processes:

Since $Q_{AB}$ and $Q_{CD}$ cancel each other out, we are left with:

Finally, let's determine the net work done and the change in internal energy. For any complete cyclic process, the system returns exactly to its initial state. Because internal energy is a state function, its net change over the entire cycle is exactly zero!

By the First Law of Thermodynamics ($Q_{\text{net}}=W_{\text{net}}+\Delta U_{\text{net}}$), if the change in internal energy is zero, the net work done must perfectly equal the net heat exchanged.

And there we have it! By breaking the cycle down into its constituent parts and applying fundamental thermodynamic principles, we've elegantly arrived at our final answers.