Animated Solution for Physics - Thermodynamics: The given $p-V$ diagram represents the thermodynamic cycle of an engine, operating with an ideal monatomic gas. The amount of heat, extracted from the source in a single cycle is

Understanding the $p-V$ Diagram

Imagine you are looking at the heartbeat of a thermodynamic engine. The $p-V$ (pressure-volume) diagram provided in the problem is exactly that—a visual representation of how an ideal monatomic gas behaves as it goes through a complete cycle. The cycle consists of four distinct paths forming a neat rectangle: an isochoric (constant volume) pressure increase, an isobaric (constant pressure) expansion, an isochoric pressure drop, and finally, an isobaric compression back to the starting state.

Our mission is to find the total amount of heat extracted from the source. But what does that actually mean in physical terms?

In the language of thermodynamics, a "source" is a hot reservoir that supplies heat to the engine. When the problem asks for the heat extracted from the source, it is asking us to calculate the total heat absorbed by the gas during the cycle ($Q>0$).

We know from the First Law of Thermodynamics and the general behavior of ideal gases that heat is typically absorbed when the temperature of the gas increases. Conversely, when the temperature drops, the gas rejects heat to the "sink". Therefore, our first strategic move is to map out the temperature at every corner of our rectangular cycle to see exactly where the gas is heating up.

The Ideal Gas Law in Action

Let's label the four corners of our cycle as states $A$, $B$, $C$, and $D$, starting from the bottom-left and moving clockwise.

Using the ideal gas equation, $pV=nRT$, we can express the temperature at any state as $T=\frac{pV}{nR}$. Let's define a base temperature $T_0$ at our starting state $A$:

Now, let's travel around the cycle: - At state $B$, the pressure is doubled to $2p_0$ while volume remains $V_0$. Thus, $T_B=\frac{(2p_0)V_0}{nR}=2T_0$. - At state $C$, both pressure and volume are doubled ($2p_0$ and $2V_0$). Thus, $T_C=\frac{(2p_0)(2V_0)}{nR}=4T_0$. - At state $D$, the pressure is back to $p_0$ but volume is $2V_0$. Thus, $T_D=\frac{p_0(2V_0)}{nR}=2T_0$.

Look closely at the temperature transitions we just calculated.

From $A\to B$, the temperature rises from $T_0$ to $2T_0$. From $B\to C$, it rises further from $2T_0$ to $4T_0$. These are our heat absorption phases!

In contrast, from $C\to D$ and $D\to A$, the temperature drops, meaning the gas is rejecting heat. Since we only care about the heat extracted from the source, we will exclusively focus on calculating the heat added during processes $A\to B$ and $B\to C$.

Process $A\to B$ is a vertical line on the $p-V$ diagram, meaning the volume is constant. This is an isochoric heating process.

For an ideal monatomic gas, the molar heat capacity at constant volume is $C_V=\frac{3}{2}R$. The heat absorbed is given by:

Substituting our values:

Since $nR{T}_0=p_0{V}_0$, we can rewrite this as:

Next, the gas moves from $B\to C$ along a horizontal line, meaning the pressure is constant. This is an isobaric expansion.

For a monatomic gas, the molar heat capacity at constant pressure is $C_p=\frac{5}{2}R$. The heat absorbed here is:

Substituting our temperature values:

Again, replacing $nR{T}_0$ with $p_0{V}_0$:

Bringing It All Together

We have successfully isolated the two phases where the engine drinks in heat from the source. To find the total heat extracted, we simply sum them up:

To add these, find a common denominator:

And there we have it! The total heat extracted from the source in a single cycle is $\frac{13}{2}{p}_0{V}_0$.

Work and Efficiency

This problem is a beautiful exercise in applying the First Law of Thermodynamics to specific gas processes. But don't stop here! As a challenge, try calculating the net work done by the engine. (Hint: It's just the area enclosed by the rectangle on the $p-V$ diagram, which is $p_0{V}_0$).

Once you have the work done and the total heat input, you can easily calculate the thermal efficiency of this engine using $\eta =\frac{W}{Q_{\text{in}}}$. Mastering these interconnected concepts is the key to conquering thermodynamics in JEE!