A sealed rigid reaction flask contains a sample of argon gas at a pressure of 120\text{ kP… — JAMB UTME

A sealed rigid reaction flask contains a sample of argon gas at a pressure of $120\text{ kPa}$ when the temperature is $30^\circ\text{C}$ . If the flask is heated to a temperature of $333^\circ\text{C}$ while maintaining a constant volume, what is the final pressure of the gas?

A
$60\text{ kPa}$
$240\text{ kPa}$ Answer
C
$1332\text{ kPa}$
D
$132\text{ kPa}$

Answer

The final pressure of the gas is

240\text{ kPa}

According to the Pressure Law (Gay-Lussac's Law), the pressure of a fixed mass of gas is directly proportional to its absolute temperature when volume remains constant (

P_1/T_1 = P_2/T_2

). Converting temperatures to Kelvin yields

T_1 = 30 + 273 = 303\text{ K}

and

T_2 = 333 + 273 = 606\text{ K}

. Substituting these into

P_2 = 120 \times (606 / 303)

gives

240\text{ kPa}

Step-by-Step Solution

Convert given temperatures from Celsius to Kelvin.

T_1 = 30 + 273 = 303\text{ K}

and

T_2 = 333 + 273 = 606\text{ K}

Gas law equations require absolute temperature measured on the Kelvin scale.

State the Pressure Law formula relating initial and final values at constant volume.

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

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

Substitute the values into the equation to calculate final pressure

P_2

P_2 = 120\text{ kPa} \times \frac{606\text{ K}}{303\text{ K}} = 120\text{ kPa} \times 2 = 240\text{ kPa}

Since absolute temperature doubles from

303\text{ K}

606\text{ K}

, the pressure must also double.

Key Concept

Pressure Law (Gay-Lussac's Law)

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