Question

An ideal gas is taken from the state A(pressure $P_0$, volume $V_0$) to the state B (pressure $P_0/2$, volume $2V_0$) along a straight line path in the P-V diagram. Select the correct statement(s) from the following.

• A. The work done by the gas in the process A to B exceeds the work that would be done by it if the system were taken from A to B along the isotherm
• B. In the T-V diagram, the path AB becomes a part of the parabola
• C. In the P-T diagram, the path AB becomes a part of hyperbola
• D. In going from A to B, the temperature T of the gas first increases to maximum value and then decreases

Answer 1

Answer: (A), (B), (D)

The correct options are
A The work done by the gas in the process A to B exceeds the work that would be done by it if the system were taken from A to B along the isotherm
B In the T-V diagram, the path AB becomes a part of the parabola
D In going from A to B, the temperature T of the gas first increases to maximum value and then decreases

Work done by the gas $=$ area under the graph on volume axis.

Clearly, from the graph, we can see that area under the isotherm is less than the area under straight line AB process.

So, option (A) is correct.

Equation of AB : $P=\left(\frac{-P_0}{2V_0}\right)V+\frac{3P_0}{2}$

$\Rightarrow \frac{nRT}{V}-\left(\frac{-P_0}{2V_0}\right)V+\frac{3P_0}{2}$ [ Applying $PV=nRT$ ]

$\Rightarrow T=\left(\frac{-P_0}{2nR{V}_0}\right)V^2+\frac{3P_0}{2}V\equiv y=-A{x}^2+Bx$

On drawing various isotherms across AB, we can see that temperature first increases then decreases.

Hence, this question has multiple answers.