A high-altitude weather research probe contains 0.030\text{ m}^3 of helium gas at a pressu… — JAMB UTME

A high-altitude weather research probe contains $0.030\text{ m}^3$ of helium gas at a pressure of $1.00 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ at ground level. During ascent, the gas expands to a volume of $0.050\text{ m}^3$ while its temperature rises to $127^\circ\text{C}$ . Assuming the helium behaves as an ideal gas, what is the new pressure of the gas in the probe?

A
$4.50 \times 10^4\text{ Pa}$
$8.00 \times 10^4\text{ Pa}$ Answer
C
$2.22 \times 10^5\text{ Pa}$
D
$2.82 \times 10^5\text{ Pa}$

Answer

The new pressure of the helium gas is

8.00 \times 10^4\text{ Pa}

Converting temperatures to absolute scale gives

T_1 = 300\text{ K}

and

T_2 = 400\text{ K}

. Rearranging the combined gas equation

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

for final pressure

P_2

yields

P_2 = 1.00 \times 10^5 \times \frac{0.030}{0.050} \times \frac{400}{300} = 8.00 \times 10^4\text{ Pa}

Step-by-Step Solution

Convert temperatures from Celsius to Kelvin

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

and

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

Gas laws require absolute temperature in Kelvin to maintain proportional thermodynamic relationships.

Apply the combined gas law formula

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \implies P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1}

Mass of the gas is constant while pressure, volume, and temperature all change simultaneously.

Substitute given values and calculate final pressure

P_2 = 1.00 \times 10^5 \times \frac{0.030}{0.050} \times \frac{400}{300} = 8.00 \times 10^4\text{ Pa}

Multiplying initial pressure by the volume contraction ratio (

0.60

) and temperature expansion ratio (

1.333

) gives the new pressure.

Key Concept

Combined Gas Law

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