A flexible research balloon is filled with 1.50\text{ m}^3 of helium gas at a temperature … — JAMB UTME

A flexible research balloon is filled with $1.50\text{ m}^3$ of helium gas at a 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 new volume of the balloon?

$2.00\text{ m}^3$ Answer
B
$7.06\text{ m}^3$
C
$1.13\text{ m}^3$
D
$1.83\text{ m}^3$

Answer

The new volume of the balloon is

2.00\text{ m}^3

The answer of

2.00\text{ m}^3

correctly uses Charles's Law with absolute temperatures converted to Kelvin (

300\text{ K}

and

400\text{ K}

), showing that heating the gas causes a proportional volume expansion.

Step-by-Step Solution

Convert temperatures from Celsius to the Kelvin absolute scale.

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

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

Gas laws strictly require thermodynamic (absolute) temperature measured in Kelvin.

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}

Volume is directly proportional to absolute temperature when pressure remains constant.

Substitute the known values and calculate

V_2

V_2 = 1.50\text{ m}^3 \times \frac{400\text{ K}}{300\text{ K}} = 2.00\text{ m}^3

Simplifying the fraction

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

gives

1.50 \times \frac{4}{3} = 2.00\text{ m}^3

Key Concept

Charles's Law (

V_1/T_1 = V_2/T_2

at constant pressure)

Estimated Time:1m 15s

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