JEE Advanced 2023 — Paper 2 — PHY-III solution

JEE Advanced2023Paper 2PHY-III

One mole of an ideal gas undergoes two different cyclic processes I and II, as shown in the $P$ - $V$ diagrams below. In cycle I, processes $a, b, c$ and $d$ are isobaric, isothermal, isobaric and isochoric, respectively. In cycle II, processes $a^{'}, b^{'}, c^{'}$ and $d^{'}$ are isothermal, isochoric, isobaric and isochoric, respectively. The total work done during cycle I is $W_{I}$ and that during cycle II is $W_{I I}$ . The ratio $W_{I} / W_{I I}$ is _______.

Your answer

Solution

Cycle I. Compute the work on each leg.

Process $a$ (isobaric expansion at $4 P_{0}$ from $V_{0}$ to $2 V_{0}$ ): $W_{a} = 4 P_{0} V_{0}$ .

Process $b$ (isothermal from $(2 V_{0}, 4 P_{0})$ to $(4 V_{0}, 2 P_{0})$ , with $n R T = 8 P_{0} V_{0}$ ): $W_{b} = 8 P_{0} V_{0} ln 2$ .

Process $c$ (isobaric compression at $2 P_{0}$ from $4 V_{0}$ to $V_{0}$ ): $W_{c} = 2 P_{0} (V_{0} - 4 V_{0}) = - 6 P_{0} V_{0}$ .

Process $d$ (isochoric at $V_{0}$ ): $W_{d} = 0$ .

$W_{I} = 4 P_{0} V_{0} + 8 P_{0} V_{0} ln 2 - 6 P_{0} V_{0} = (8 ln 2 - 2) P_{0} V_{0}$ .

Cycle II.

Process $a^{'}$ (isothermal from $(V_{0}, 4 P_{0})$ to $(2 V_{0}, 2 P_{0})$ , with $n R T = 4 P_{0} V_{0}$ ): $W_{a^{'}} = 4 P_{0} V_{0} ln 2$ .

Process $b^{'}$ (isochoric at $2 V_{0}$ ): $W_{b^{'}} = 0$ .

Process $c^{'}$ (isobaric compression at $P_{0}$ from $2 V_{0}$ to $V_{0}$ ): $W_{c^{'}} = P_{0} (V_{0} - 2 V_{0}) = - P_{0} V_{0}$ .

Process $d^{'}$ (isochoric at $V_{0}$ ): $W_{d^{'}} = 0$ .

$W_{I I} = 4 P_{0} V_{0} ln 2 - P_{0} V_{0} = (4 ln 2 - 1) P_{0} V_{0}$ .

Ratio:

$\frac{W _{I}}{W _{I I}} = \frac{8 ln 2 - 2}{4 ln 2 - 1} = \frac{2 ( 4 ln 2 - 1 )}{4 ln 2 - 1} = 2$ .

← Back to full paper

Practice more PHY-III

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →