Animated Solution for Physics - Thermodynamics: One mole of a monoatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature (V-T) diagram. The correct statement(s) is/are: [R is the gas constant]

Imagine you are sitting in the JEE Advanced exam hall. The clock is ticking. You look at this V-T graph, start calculating the areas and heat transfers, and suddenly... panic sets in. The numbers aren't adding up! None of the options seem to match your rigorous calculations.

This is a classic moment where elite students separate themselves from the rest. They don't panic; they look for the hidden symmetry. Let's embark on a thrilling journey through this thermodynamic cycle and uncover the famous "trap" of the JEE Advanced 2019 paper.

Our first mission is to translate the visual geometry of the V-T graph into physical thermodynamic processes.

Look at the line representing process $1\to 2$. It is a straight line that, if extended backwards, would pass perfectly through the origin. Mathematically, this means that Volume is directly proportional to Temperature ($V\propto T$). According to the Ideal Gas Law ($PV=nRT$), this proportionality only holds true when the pressure $P$ is constant. Therefore, process $1\to 2$ is an isobaric expansion.

For an isobaric process, the heat exchanged is given by $Q=n{C}_P\Delta T$. Since we are dealing with a monoatomic gas, $C_P=\frac{5}{2}R$.

Next, process $2\to 3$ is a perfectly horizontal line. The volume is locked at $2V_0$. This is an isochoric process, meaning the gas does absolutely zero work ($W=0$).

A Flawed Diagram

Here is where the plot thickens. If you look closely at the graph, the dashed line from point 3 drops down to a temperature of $\frac{3T_0}{2}$. If you blindly trust this label and proceed to calculate the heat transfers and work done, you will hit a dead end. The ratios will not match any of the given options.

Why? Because the diagram contains a flaw! The intended problem was a beautifully symmetric, standard rectangular cycle in the P-V plane. A P-V rectangle consists of alternating isobaric and isochoric processes.

If the cycle is indeed a P-V rectangle, then process $3\to 4$ must also be isobaric. For a line to be isobaric on a V-T graph, it must pass through the origin. If you draw a straight line from the origin through point 4 $(T_0/2,V_0)$ and extend it to a volume of $2V_0$, it will hit exactly at temperature $T_0$.

Therefore, the true, intended temperature for point 3 is $T_0$, not $\frac{3T_0}{2}$. Sometimes, you have to look past the typo and see the physics the examiner intended!

Armed with the corrected temperature $T_3=T_0$, the math flows like butter. Let's calculate the heat rejected during the isochoric process $2\to 3$. We use $C_V=\frac{3}{2}R$ for a monoatomic gas.

Now, let's check Option (B) by finding the magnitude of the ratio of these heat transfers:

This perfectly matches Option (B)!

To check Option (A), we need the total net work done by the gas. We could painstakingly calculate the work for each individual process and sum them up, but there is a much more elegant way.

Remember our intended P-V graph? It's a perfect rectangle! The net work done in any cyclic process is simply the area enclosed by the cycle on a P-V diagram.

Let's find the dimensions of this rectangle. - The width (change in volume) is $2V_0-V_0=V_0$. - The height (change in pressure) is $P_1-P_4$.

We know $P_1=\frac{R{T}_0}{V_0}=P_0$. For point 4, $P_4=\frac{R(T_0/2)}{V_0}=\frac{P_0}{2}$.

So, the height of the rectangle is $P_0-\frac{P_0}{2}=\frac{P_0}{2}$.

The area of the rectangle is:

Using the ideal gas law ($P_0{V}_0=R{T}_0$), we get:

This elegantly proves that Option (A) is absolutely correct. See how powerful a simple P-V transformation can be? By trusting the symmetry of physics over a flawed label, a seemingly impossible problem becomes a masterpiece of logic.