A sealed rigid glass bulb contains a fixed mass of helium gas at an initial pressure of 1.… — JAMB UTME

A sealed rigid glass bulb contains a fixed mass of helium gas at an initial pressure of $1.20 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . Assuming the volume of the bulb remains constant, calculate the final pressure of the gas in $\text{Pa}$ when it is heated to $127^\circ\text{C}$ .

Answer: 160000 Pa

Answer

The final pressure of the gas is

1.60 \times 10^5\text{ Pa}

(

160\,000\text{ Pa}

The correct result of

160\,000\text{ Pa}

(

1.60 \times 10^5\text{ Pa}

) is obtained by applying the Pressure Law

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

using absolute temperatures (

T_1 = 300\text{ K}

T_2 = 400\text{ K}

Step-by-Step Solution

Convert the initial and final temperatures from degrees Celsius to the absolute temperature scale (Kelvin).

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

and

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

All gas law equations require absolute temperatures in Kelvin.

Apply the Pressure Law (Gay-Lussac's Law) for a fixed mass of gas at constant volume.

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

The pressure of a gas is directly proportional to its absolute temperature when volume remains constant.

Substitute the known values to find

P_2

P_2 = 1.20 \times 10^5\text{ Pa} \times \frac{400\text{ K}}{300\text{ K}} = 1.60 \times 10^5\text{ Pa} = 160\,000\text{ Pa}

Multiplying

1.20 \times 10^5

by the temperature ratio

\frac{4}{3}

gives

1.60 \times 10^5\text{ Pa}

Key Concept

Pressure Law (Gay-Lussac's Law) and absolute temperature conversion

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