A rigid scuba diving cylinder contains a fixed mass of compressed air at an initial absolu… — JAMB UTME

A rigid scuba diving cylinder contains a fixed mass of compressed air at an initial absolute pressure of $2.00 \times 10^7\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . The cylinder is left under direct sunlight on a boat deck, causing its temperature to rise to $77^\circ\text{C}$ . Assuming the volume of the cylinder remains constant, what is the new pressure of the air inside the cylinder?

$2.33 \times 10^7\text{ Pa}$ Answer
B
$5.70 \times 10^7\text{ Pa}$
C
$1.71 \times 10^7\text{ Pa}$
D
$5.13 \times 10^6\text{ Pa}$

Answer

The final pressure of the air inside the cylinder is

2.33 \times 10^7\text{ Pa}

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

P \propto T

). Converting the given temperatures yields

T_1 = 300\text{ K}

and

T_2 = 350\text{ K}

. Substituting these into

P_2 = P_1(T_2 / T_1)

gives

2.00 \times 10^7 \times (350 / 300) = 2.33 \times 10^7\text{ Pa}

Step-by-Step Solution

Convert given temperatures from Celsius to Kelvin.

T_1 = 27^\circ\text{C} + 273 = 300\text{ K}

, and

T_2 = 77^\circ\text{C} + 273 = 350\text{ K}

All gas laws strictly require absolute temperature measured in Kelvin.

State and rearrange 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}

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

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

P_2 = 2.00 \times 10^7\text{ Pa} \times \frac{350\text{ K}}{300\text{ K}} \approx 2.33 \times 10^7\text{ Pa}

Evaluates the final pressure accurately after heating.

Key Concept

Pressure Law (Gay-Lussac's Law) for Ideal Gases

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