Question

An ideal gas is taken from the state A (pressure p, volume V) to the state B (pressure p /2, volume 2 V) along a straight line path in the p-V diagram. Select the correct statements 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 an isotherm
• B. In the T-V diagram, the path AB becomes a part of a parabola
• C. In the p-T diagram, the path AB becomes a part of a hyperbola
• D. In going from A to B, the temperature T of the gas first increases to a maximum value and then decreases

Answer 1

Answer: (D)

In going from A to B, the temperature T of the gas first increases to a maximum value and then decreases

Answer 2

(a) Work done = Area under p- V graph
\hspace20mm A_1 > A_2
$\therefore \, \, \, \, \, \, \, \, \, W_{given \, process > W_{isotermal \, process }}$
(b) In the given process p_V equation will be of a straight line with negative slope and positive intercept i.e.,
$p=-\alpha V +\beta (Here \, \alpha \, and \, beta \, are \, positive \, constants)$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, p_V =- \alpha V^2 +\beta V$
$\Rightarrow \, \, \, \, \, \, \, \, nRT =-\alpha V^2 +\beta V$
$\Rightarrow \, \, \, \, \, \, \, T=\frac{1}{nR}(-\alpha ^2 +\beta V) \, \, \, \, \, \,$...(i)
This is an equation of parabola in T and V
(d) $\frac{dT}{dV}=0-\beta -2\alpha V \, \, \, \Rightarrow \, \, \, V=\frac{B}{2\alpha}$
Now, $\, \, \, \, \, \, \frac{d^2 T}{dV^2 =-2\alpha =-ve}$
ie, T has some maximum value.
Now, $T ? pV \, and \, (pA)_A =(pV)_B$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, T_A =T_B$
We conclude that temperatures are same at A and B and in between temperature has a maximum value.
Therefore, in going from A to B, T will first increase to a maximum value and then decrease.