A rigid steel cylinder contains methane gas at a pressure of 150\text{ kPa} and a temperat… — JAMB UTME

A rigid steel cylinder contains methane gas at a pressure of $150\text{ kPa}$ and a temperature of $47^\circ\text{C}$ . If the cylinder is heated to $127^\circ\text{C}$ while maintaining a constant volume, what is the final pressure of the gas in $\text{kPa}$ ?

Answer: 187.5 kPa

Answer

The final pressure of the methane gas is

187.5\text{ kPa}

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

P \propto T

). Converting temperatures to Kelvin gives

T_1 = 320\text{ K}

and

T_2 = 400\text{ K}

. Applying

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

yields

\frac{150}{320} = \frac{P_2}{400}

, which gives

P_2 = 187.5\text{ kPa}

Step-by-Step Solution

Convert given temperatures to the Kelvin absolute scale.

T_1 = 47 + 273 = 320\text{ K}

and

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

All gas law calculations require temperatures to be expressed in Kelvin.

Set up the Pressure Law equation relating pressure and absolute temperature at constant volume.

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

According to Gay-Lussac's Law, pressure is directly proportional to absolute temperature when volume is fixed.

Substitute the known values into the equation and solve for

P_2

P_2 = \frac{150 \times 400}{320} = 187.5\text{ kPa}

Cross-multiplication gives

P_2 = 187.5\text{ kPa}

Key Concept

Pressure Law (Gay-Lussac's Law)

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic