A flexible gas storage container at a research laboratory holds 0.30\text{ m}^3 of helium … — JAMB UTME

A flexible gas storage container at a research laboratory holds $0.30\text{ m}^3$ of helium gas at an initial temperature of $27^\circ\text{C}$ . If the gas is heated at constant pressure until its temperature reaches $127^\circ\text{C}$ , what is the final volume occupied by the gas?

$0.40\text{ m}^3$ Answer
B
$1.41\text{ m}^3$
C
$0.23\text{ m}^3$
D
$0.10\text{ m}^3$

Answer

The final volume of the gas is

0.40\text{ m}^3

According to Charles's law, at constant pressure, the volume of a gas is directly proportional to its absolute temperature (

V \propto T

). Converting

27^\circ\text{C}

300\text{ K}

and

127^\circ\text{C}

400\text{ K}

, the volume expands by a ratio of

\frac{400}{300} = \frac{4}{3}

. Multiplying the initial volume

0.30\text{ m}^3

\frac{4}{3}

gives

0.40\text{ m}^3

Step-by-Step Solution

Convert initial and final temperatures from Celsius to Kelvin

T_1 = 27 + 273 = 300\text{ K}

T_2 = 127 + 273 = 400\text{ K}

Gas laws require absolute temperatures in Kelvin to maintain proportional relationships.

Apply Charles's Law for constant pressure processes

\frac{V_1}{T_1} = \frac{V_2}{T_2} \implies V_2 = V_1 \times \frac{T_2}{T_1}

At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.

Substitute the known values and calculate the final volume

V_2 = 0.30\text{ m}^3 \times \frac{400\text{ K}}{300\text{ K}} = 0.40\text{ m}^3

Performing arithmetic yields the expanded volume.

Key Concept

Charles's Law

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