An aerosol canister containing an ideal gas at an initial pressure of 2.20 \times 10^5\tex… — JAMB UTME

An aerosol canister containing an ideal gas at an initial pressure of $2.20 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ is accidentally thrown into a fire, causing the temperature of the gas to rise to $327^\circ\text{C}$ . Assuming the volume of the canister remains constant, what is the new pressure of the gas inside the canister?

A
$1.10 \times 10^5\text{ Pa}$
B
$2.66 \times 10^6\text{ Pa}$
$4.40 \times 10^5\text{ Pa}$ Answer
D
$2.45 \times 10^5\text{ Pa}$

Answer

The new pressure of the gas inside the canister is

4.40 \times 10^5\text{ Pa}

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

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

). Converting the temperatures gives

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

and

T_2 = 327 + 273 = 600\text{ K}

. Multiplying the initial pressure by the ratio

\frac{600\text{ K}}{300\text{ K}} = 2

gives

4.40 \times 10^5\text{ Pa}

Step-by-Step Solution

Convert initial and final temperatures from degrees Celsius to Kelvin.

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

and

T_2 = 327^\circ\text{C} + 273 = 600\text{ K}

All calculations using gas laws require absolute temperature measured in Kelvin.

State and rearrange Pressure's 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 to calculate the final pressure

P_2

P_2 = 2.20 \times 10^5\text{ Pa} \times \frac{600\text{ K}}{300\text{ K}} = 4.40 \times 10^5\text{ Pa}

Since absolute temperature doubles from

300\text{ K}

600\text{ K}

, the final pressure must also double.

Key Concept

Pressure Law (Gay-Lussac's Law) and Absolute Temperature Conversion

Estimated Time:1m 30s

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