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MHTCET 20 Apr 2026 Shift 2 Paper

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Question : 17 of 150

Marks: +1, -0

An ideal gas having pressure

P

V

and temperature

T

is expanded isothermally, to a volume

3 V

and final pressure

P_{I}

. The same gas is expanded adiabatically to a volume

3 V

, the final pressure being

P_{A}

\frac{P_{A}}{P_{I}}

\left(\frac{C_{P}}{C_{V}}=\gamma\right)

$3 \gamma$
$3^{\gamma}$
$3^{(1-\gamma)}$
$3^{(\gamma-1)}$

Solution:

Concept:

For an isothermal process,

PV

is constant; for an adiabatic process,

PV^{\gamma}

is constant.

Explanation:

The gas expands isothermally from volume

V

to

3V

.

Using

PV = \text{constant}

:

P V = P_I (3V)

Therefore, the final isothermal pressure is:

P_I = \frac{P}{3}

Now, for the adiabatic expansion from volume

V

to

3V

:

Using

PV^{\gamma} = \text{constant}

:

P V^{\gamma} = P_A (3V)^{\gamma}

P_A = \frac{P V^{\gamma}}{3^{\gamma} V^{\gamma}}

Thus, the final adiabatic pressure is:

P_A = \frac{P}{3^{\gamma}}

The required ratio is:

\frac{P_A}{P_I} = \frac{\frac{P}{3^{\gamma}}}{\frac{P}{3}}

\frac{P_A}{P_I} = \frac{3}{3^{\gamma}} = 3^{1-\gamma}

Answer:

\frac{P_A}{P_I} = 3^{(1-\gamma)}

Hence, the correct option is C.

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