A rigid steel cylinder contains a gas at a pressure of 1.20\text{ atm} and a temperature o… — JAMB UTME

A rigid steel cylinder contains a gas at a pressure of $1.20\text{ atm}$ and a temperature of $27^\circ\text{C}$ . If the cylinder is heated to $127^\circ\text{C}$ while keeping the volume constant, what is the final pressure of the gas?

$1.60\text{ atm}$ Answer
B
$5.64\text{ atm}$
C
$0.90\text{ atm}$
D
$2.20\text{ atm}$

Answer

The final pressure of the gas inside the cylinder is

1.60\text{ atm}

According to the Pressure Law, for a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature (

P_1/T_1 = P_2/T_2

). Converting temperatures to Kelvin gives

T_1 = 300\text{ K}

and

T_2 = 400\text{ K}

. Solving for

P_2

yields

1.20 \times (400/300) = 1.60\text{ atm}

Step-by-Step Solution

Convert initial and final temperatures from degrees Celsius to Kelvin.

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

and

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

Gas law equations require absolute temperatures in Kelvin.

Apply the 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}

Pressure is directly proportional to absolute temperature when volume is constant.

Substitute the known values into the equation and calculate the final pressure.

P_2 = 1.20\text{ atm} \times \frac{400\text{ K}}{300\text{ K}} = 1.60\text{ atm}

Simplifying

\frac{400}{300}

\frac{4}{3}

gives

1.20 \times \frac{4}{3} = 1.60\text{ atm}

Key Concept

Pressure-Temperature relationship (Pressure Law) and conversion to absolute temperature scale.

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