An aerosol spray can contains a fixed mass of gas at a pressure of 100\text{ kPa} and a te… — JAMB UTME

An aerosol spray can contains a fixed mass of gas at a pressure of $100\text{ kPa}$ and a temperature of $17^\circ\text{C}$ . If the volume of the container remains constant, what is the pressure inside the can when its temperature increases to $307^\circ\text{C}$ ?

$200\text{ kPa}$ Answer
B
$106\text{ kPa}$
C
$1806\text{ kPa}$
D
$50\text{ kPa}$

Answer

The pressure inside the can when heated to

307^\circ\text{C}

200\text{ kPa}

The correct answer is

200\text{ kPa}

. According to Gay-Lussac's Pressure Law, for a gas at constant volume, pressure is directly proportional to absolute temperature in Kelvin. Converting both temperatures to Kelvin gives

T_1 = 290\text{ K}

and

T_2 = 580\text{ K}

. Since absolute temperature doubles, the pressure also doubles from

100\text{ kPa}

200\text{ kPa}

Step-by-Step Solution

Convert given temperatures from Celsius to Kelvin.

T_1 = 17^\circ\text{C} + 273 = 290\text{ K}

and

T_2 = 307^\circ\text{C} + 273 = 580\text{ K}

Gas law equations require absolute temperatures in Kelvin.

Apply Gay-Lussac's Pressure Law formula at constant volume.

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

Pressure of a fixed mass of gas is directly proportional to its absolute temperature.

Substitute the values and calculate the final pressure

P_2

P_2 = 100\text{ kPa} \times \frac{580\text{ K}}{290\text{ K}} = 100\text{ kPa} \times 2 = 200\text{ kPa}

Multiplying the initial pressure by the ratio of absolute temperatures gives the final pressure.

Key Concept

Pressure Law (Gay-Lussac's Law of Temperature-Pressure)

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