A rigid steel container holds a fixed mass of gas at an initial pressure of 1.20 \times 10… — JAMB UTME

A rigid steel container holds a fixed mass of gas at an initial pressure of $1.20 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . If the temperature of the gas is increased to $127^\circ\text{C}$ while keeping its volume constant, what is the final pressure of the gas?

$1.60 \times 10^5\text{ Pa}$ Answer
B
$5.64 \times 10^5\text{ Pa}$
C
$0.90 \times 10^5\text{ Pa}$
D
$4.44 \times 10^5\text{ Pa}$

Answer

The final pressure of the gas is

1.60 \times 10^5\text{ Pa}

According to Gay-Lussac's law, at constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature (

P_1/T_1 = P_2/T_2

). Converting the temperatures to Kelvin gives

T_1 = 300\text{ K}

and

T_2 = 400\text{ K}

. Solving for

P_2

yields

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

Step-by-Step Solution

Convert temperatures from Celsius to Kelvin

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

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

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

Apply Pressure Law (Gay-Lussac's Law) for constant volume

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

For a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature.

Substitute the known values to calculate the final pressure

P_2 = (1.20 \times 10^5\text{ Pa}) \times \frac{400\text{ K}}{300\text{ K}} = 1.60 \times 10^5\text{ Pa}

Multiplying the initial pressure by the temperature expansion factor yields the final pressure.

Key Concept

Gay-Lussac's Law (Pressure Law)

Question

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