A fixed mass of gas is sealed inside a rigid vessel. Complete the following statement by c… — JAMB UTME

A fixed mass of gas is sealed inside a rigid vessel. Complete the following statement by calculating the final pressure after temperature change.

Answer:A gas sealed in a rigid container exerts a pressure of

240\text{ kPa}

127^\circ\text{C}

. When cooled at constant volume to

-73^\circ\text{C}

, the final pressure of the gas is 【120】

\text{kPa}

120

According to Gay-Lussac's Pressure Law, for a given mass of gas at constant volume, pressure is directly proportional to absolute temperature (

P \propto T

). Converting temperatures to Kelvin gives

T_1 = 400\text{ K}

and

T_2 = 200\text{ K}

. Substituting into

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

yields

P_2 = 240 \times \frac{200}{400} = 120\text{ kPa}

Convert initial and final temperatures from Celsius to Kelvin using

T = t + 273

T_1 = 127 + 273 = 400\text{ K}

and

T_2 = -73 + 273 = 200\text{ K}

Gas law calculations require thermodynamic absolute temperature in Kelvin.

Apply Pressure Law formula

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

at constant volume.

\frac{240}{400} = \frac{P_2}{200}

Pressure of a fixed mass of gas is directly proportional to absolute temperature when volume remains constant.

Solve for final pressure

P_2

P_2 = 240 \times \frac{200}{400} = 240 \times 0.5 = 120\text{ kPa}

Halving absolute temperature halves the pressure exerted by gas molecules.

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

Estimated Time:1m 30s

Question