A fixed mass of oxygen gas contained in a rigid metal vessel exerts a pressure of 1.20\tex… — JAMB UTME

A fixed mass of oxygen gas contained in a rigid metal vessel exerts a pressure of $1.20\text{ atm}$ at a temperature of $27^\circ\text{C}$ . If the volume of the vessel remains constant, at what absolute temperature in Kelvin (K) will the gas exert a pressure of $1.80\text{ atm}$ ?

Answer: 450 K

Answer

The final absolute temperature of the gas is

450\text{ K}

According to Gay-Lussac's Pressure Law, for a fixed mass of gas at constant volume,

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

. Converting the initial temperature

27^\circ\text{C}

to Kelvin gives

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

. Substituting

P_1 = 1.20\text{ atm}

P_2 = 1.80\text{ atm}

, and

T_1 = 300\text{ K}

gives

T_2 = \frac{1.80 \times 300}{1.20} = 450\text{ K}

Step-by-Step Solution

Convert the given initial temperature from Celsius to Kelvin.

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

Gas laws require absolute temperature in Kelvin for proportional relationship calculations.

State the Pressure Law (Gay-Lussac's Law) equation for a fixed volume of gas.

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

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

Substitute the given values into the Pressure Law equation.

\frac{1.20\text{ atm}}{300\text{ K}} = \frac{1.80\text{ atm}}{T_2}

Insert

P_1 = 1.20\text{ atm}

T_1 = 300\text{ K}

, and

P_2 = 1.80\text{ atm}

Rearrange and solve for

T_2

T_2 = \frac{1.80 \times 300}{1.20} = 450\text{ K}

Cross-multiplying yields the final absolute temperature.

Key Concept

Pressure Law (Gay-Lussac's Law)

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