Animated Solution for Physics - Thermodynamics: One mole of an ideal monoatomic gas is taken round the cyclic process ABCA as shown in figure. Calculate (a) the work done by the gas. (b) the heat rejected by the gas in the path CA and the heat absorbed by the gas in the path AB. (c) the net heat absorbed by the gas in the path BC. (d) the maximum temperature attained by the gas during the cycle.

Imagine a gas trapped in a cylinder, expanding and compressing, heating up and cooling down, only to return exactly to where it started. This is a cyclic process, the beating heart of every engine and refrigerator. In this problem, we are given a clockwise cycle ABCA for one mole of an ideal monoatomic gas on a $p-V$ diagram. Our mission is to dissect this cycle, calculating work, heat, and temperature at every step.

The Area Under the Curve

The first question asks for the net work done by the gas. In thermodynamics, the $p-V$ diagram is our best friend because the work done is simply the area under the curve. For a closed cycle, the net work done is the area enclosed by the loop.

Since our cycle ABCA forms a right-angled triangle, calculating the area is straightforward geometry. The base of the triangle is the change in volume, $2V_0-V_0=V_0$. The height is the change in pressure, $3p_0-p_0=2p_0$.

Using the formula for the area of a triangle, we get:

Because the cycle is clockwise, the expansion happens at a higher pressure than the compression, meaning the net work done by the gas is positive.

Isobaric and Isochoric Heat Exchange

Next, we need to find the heat exchanged in paths CA and AB. Let's look at path CA first. It's a horizontal line, which means the pressure is constant. This is an isobaric process. For an isobaric process, the heat exchanged is $Q=n{C}_p\Delta T$.

Using the ideal gas law, $pV=nRT$, we can rewrite $n\Delta T$ as $\frac{\Delta (pV)}{R}$. For a monoatomic gas, $C_p=\frac{5}{2}R$. Substituting the values for points A and C: The negative sign tells us that heat is rejected by the gas.

Now, let's examine path AB. It's a vertical line, meaning the volume is constant. This is an isochoric process. Here, the heat exchanged is $Q=n{C}_V\Delta T$. For a monoatomic gas, $C_V=\frac{3}{2}R$. This value is positive, meaning heat is absorbed by the gas.

The First Law and the Complete Cycle

Part (c) asks for the net heat absorbed in path BC. We could calculate it by finding the work done and change in internal energy for path BC, but there is a much more elegant way.

Let's zoom out and look at the entire cycle. Internal energy is a state function. Because the gas returns to its exact starting point, the net change in internal energy for the complete cycle is zero ($\Delta U=0$).

According to the First Law of Thermodynamics, $\Delta U=Q_{\text{net}}-W_{\text{net}}$. Therefore, for a complete cycle, the net heat absorbed equals the net work done:

We already know $Q_{AB}$ and $Q_{CA}$. Let's plug them in: This is a beautiful demonstration of how macroscopic conservation laws can save us from tedious path-specific calculations!

The Calculus of Maximum Temperature

Finally, we are asked to find the maximum temperature attained during the cycle. From the ideal gas law, $T=\frac{pV}{nR}$. Since $n=1$, temperature is directly proportional to the product $pV$.

Let's check the vertices: - At A: $T_A=\frac{p_0{V}_0}{R}$ - At B: $T_B=\frac{3p_0{V}_0}{R}$ - At C: $T_C=\frac{2p_0{V}_0}{R}$

It's tempting to say $T_B$ is the maximum. However, path BC is a straight line where both $p$ and $V$ are changing. The product $pV$ along this line forms a parabola, which might have a peak higher than the endpoints!

Let's find the equation of the line BC. It passes through $(V_0,3p_0)$ and $(2V_0,p_0)$. The slope is: Using the point-slope form, the equation of the line is:

Now, substitute this into the temperature equation: To find the maximum temperature, we take the derivative with respect to $V$ and set it to zero:

This volume is between $V_0$ and $2V_0$, so the maximum temperature indeed occurs along the path BC. Let's plug this volume back into our temperature function:

And there we have it! By combining geometry, the laws of thermodynamics, and a touch of calculus, we've completely unraveled the secrets of this cyclic process.