Two litre of an ideal gas at 42^ ^ C is heated at a constant pressure to 357^ ^ C . Then the final volume is

Correct answer: C. 4L Given, initial volume V_1 = 2L Initial temperature T_1 = 42^ ^ C = 42 + 273 = 315K Final temperature T_2 = 357^ ^ C = 357 + 273 = 630K At constant pressure, using Charles's law: (V_1)/(T_1) = (V_2)/(T_2) (2)/(315) = (V_2)/(630) V_2 = (2 × 630)/(315) = 4L Answer: 4L

MHT CET 202626 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 26 April
Shift: Morning Shift
Subject: Physics
Chapter: Kinetic Theory of Gases

MHT CET 2026 · 26 April · Morning Shift · Physics · Kinetic Theory of Gases

Two litre of an ideal gas at 42^ ^ C is heated at a constant...

Two litre of an ideal gas at

42^{}^{\circ}\text{C}

is heated at a constant pressure to

357^{}^{\circ}\text{C}

. Then the final volume is

Options

A $6\text{L}$
B $2\text{L}$
C $4\text{L}$
D $3.5\ \text{L}$

Correct answer

C. $4\text{L}$

Step-by-step solution

Given, initial volume

V_1 = 2\text{L}

Initial temperature

T_1 = 42^{}^{\circ}\text{C} = 42 + 273 = 315\text{K}

Final temperature

T_2 = 357^{}^{\circ}\text{C} = 357 + 273 = 630\text{K}

At constant pressure, using Charles's law:

\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}

\dfrac{2}{315} = \dfrac{V_2}{630}

V_2 = \dfrac{2 \times 630}{315} = 4\text{L}

Answer:

4\text{L}

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