Animated Solution for Physics - Thermodynamics: An ideal gas is taken from the state $A$ (pressure $p$, volume $V$) to the state $B$ (pressure $p/2$, volume $2V$) along a straight line path in the $p-V$ diagram. Select the correct statements from the following

Imagine a gas expanding in a cylinder, but instead of following a familiar curve like an isotherm or an adiabat, it traces a perfect straight line on the $p-V$ diagram. This is a classic polytropic process, and it hides some beautiful mathematical secrets. Let's break down the journey from state $A$ to state $B$.

The work done by a gas is geometrically represented by the area under the curve on a $p-V$ diagram. Our gas travels from $(V,p)$ to $(2V,p/2)$ along a straight line.

What if the gas had expanded isothermally between these same two volumes? An isothermal process follows the equation $pV=\text{constant}$, which graphs as a rectangular hyperbola. If you plot this hyperbola, you'll notice it sags below the straight line connecting the two states. Because the straight line lies strictly above the isotherm, the area under the straight line is greater. Therefore, the work done in the straight-line process exceeds the work done in the isothermal process.

To understand how temperature behaves, we first need the equation of our straight-line path. Since the line has a negative slope and a positive $p$-intercept, we can write it as:

where $\alpha$ and $\beta$ are positive constants. To bring temperature into the mix, we multiply the entire equation by $V$:

Now, we invoke the ideal gas law, $pV=nRT$, and substitute it in:

Look closely at this equation. It's a quadratic equation in $V$ with a negative leading coefficient ($-\alpha$). In the $T-V$ plane, this represents a downward-opening parabola!

What if we wanted to plot $p$ versus $T$? We can rearrange our straight-line equation to solve for $V$:

Substitute this back into the ideal gas law:

Once again, we have a quadratic equation, this time in terms of $p$. This means the $p-T$ graph is also a parabola, not a hyperbola.

Finally, let's evaluate the temperatures at the endpoints. At state $A$, the product of pressure and volume is simply $pV$. At state $B$, the product is $(p/2)\times (2V)$, which simplifies to $pV$.

Since the product $pV$ is identical at both states, the ideal gas law dictates that the temperatures must also be identical:

We established earlier that the $T-V$ graph is a downward-opening parabola. If a downward-opening parabola has the same height at two different points, it must rise to a peak somewhere between them. Therefore, as the gas expands from $A$ to $B$, its temperature first increases, reaches a maximum value, and then decreases back to its starting temperature.

This elegant interplay of geometry and thermodynamics perfectly validates our conclusions!