A rigid metallic cylinder contains a fixed mass of gas at an initial pressure of 1.2\text{… — JAMB UTME

A rigid metallic cylinder contains a fixed mass of gas at an initial pressure of $1.2\text{ atm}$ and a temperature of $27^\circ\text{C}$ . If the gas is heated until its temperature reaches $177^\circ\text{C}$ while the volume remains constant, what is the final pressure of the gas in atmospheres?

Answer: 1.8 atm

Answer

The final pressure of the gas is

1.8\text{ atm}

Converting the temperatures to Kelvin (

T_1 = 300\text{ K}

and

T_2 = 450\text{ K}

) and applying Gay-Lussac's Pressure Law

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

gives

P_2 = 1.2 \times \frac{450}{300} = 1.8\text{ atm}

Step-by-Step Solution

Convert given temperatures to the Kelvin scale

T_1 = 300\text{ K}

and

T_2 = 450\text{ K}

Gas laws require absolute temperature values measured in Kelvin (

T_K = t_C + 273

Set up Gay-Lussac's Pressure Law proportion

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

P_2 = P_1 \times \frac{T_2}{T_1} = 1.2 \times \frac{450}{300}

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

Perform the multiplication to determine the final pressure

P_2 = 1.8\text{ atm}

Multiplying

1.2

by the ratio

1.5

yields

1.8\text{ atm}

Key Concept

Pressure Law (Gay-Lussac's Law) states that for a given mass of gas at constant volume, the pressure is directly proportional to its absolute temperature in Kelvin (

P \propto T

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Key Concept

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