Animated Solution for Physics - Thermodynamics: Match the following for the given process

Imagine you are a detective, and the $p-V$ diagram is your crime scene. Every line, every point holds a clue about what the gas is experiencing. Is it heating up? Is it doing work? Is it absorbing heat? To crack this case, we need our master key: the First Law of Thermodynamics.

Our Guiding Light

The First Law is beautifully simple: $\Delta Q=\Delta U+\Delta W$. It tells us that the heat $\Delta Q$ supplied to a gas goes into two things: changing its internal energy $\Delta U$ and doing work $\Delta W$.

To find the sign of work $\Delta W$, we just look at the volume. If the gas expands, it pushes against its surroundings, doing positive work. If it's compressed, work is negative. If the volume doesn't change, work is zero.

For the internal energy $\Delta U$, we look at the temperature. For an ideal gas, $pV=nRT$. This means the temperature is directly proportional to the product of pressure and volume ($pV$). If $pV$ increases, the gas is heating up, and $\Delta U$ is positive. If $pV$ decreases, it's cooling down, and $\Delta U$ is negative.

Let's start with the vertical paths, where the volume is constant. These are called isochoric processes.

Process $J\to K$: The gas goes from a pressure of $30$ to $10$ while the volume stays locked at $10{\text{ m}}^3$. Because the volume doesn't budge, the gas does no work: $\Delta W=0$. What about the temperature? At $J$, $pV=30\times 10=300$. At $K$, $pV=10\times 10=100$. The product $pV$ has plummeted, meaning the gas has cooled down significantly. Therefore, $\Delta U<0$. Plugging this into our First Law: $\Delta Q=\text{negative}+0$, which means $\Delta Q<0$. The gas is losing heat! This matches option (s).

Process $L\to M$: This is the other vertical path. The volume is constant at $20{\text{ m}}^3$, so again, $\Delta W=0$. This time, the pressure increases from $10$ to $20$. The product $pV$ jumps from $200$ (at $L$) to $400$ (at $M$). The gas is heating up, so $\Delta U>0$. Our First Law tells us: $\Delta Q=\text{positive}+0$, meaning $\Delta Q>0$. The gas is absorbing heat. This matches option (p).

Now let's look at the horizontal path, Process $K\to L$. Here, the pressure is constant at $10$, but the volume expands from $10$ to $20{\text{ m}}^3$. Because the gas is expanding, it's doing positive work: $\Delta W>0$. Let's check the temperature. The product $pV$ goes from $100$ (at $K$) to $200$ (at $L$). The temperature is rising, so $\Delta U>0$. With both work and internal energy being positive, their sum must be positive: $\Delta Q>0$. The gas is absorbing heat while doing work. This matches options (p) and (r).

Finally, we have the slanted path, Process $M\to J$. The volume is being crushed from $20$ down to $10{\text{ m}}^3$. Because it's a compression, the work done is negative: $\Delta W<0$. What about the internal energy? At $M$, $pV=400$. At $J$, $pV=300$. The product $pV$ is decreasing, which means the gas is cooling down, so $\Delta U<0$. Both terms in our First Law are negative, so their sum must be negative: $\Delta Q<0$. The gas is losing heat while being compressed. This matches options (q) and (s).

By breaking down the cycle into individual processes and applying the First Law, we've solved the mystery. We didn't need complex integrals or messy algebra—just a solid understanding of how pressure, volume, and temperature interact. This is the true elegance of thermodynamics!