An industrial compression chamber fitted with a frictionless piston contains 0.25\text{ m}… — JAMB UTME

An industrial compression chamber fitted with a frictionless piston contains $0.25\text{ m}^3$ of an ideal gas at an initial pressure of $1.6 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . The gas expands while being heated to a final temperature of $87^\circ\text{C}$ until its pressure drops to $8.0 \times 10^4\text{ Pa}$ . What is the final volume occupied by the gas?

A
$0.15\text{ m}^3$
B
$0.42\text{ m}^3$
$0.60\text{ m}^3$ Answer
D
$1.61\text{ m}^3$

Answer

0.60\text{ m}^3

The combined gas law states that

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

. Converting the temperatures to Kelvin gives

T_1 = 300\text{ K}

and

T_2 = 360\text{ K}

. Rearranging for

V_2

gives

V_2 = 0.25 \times \left(\frac{1.6 \times 10^5}{8.0 \times 10^4}\right) \times \left(\frac{360}{300}\right) = 0.25 \times 2.0 \times 1.2 = 0.60\text{ m}^3

Step-by-Step Solution

Convert all given temperatures from degrees Celsius to Kelvin.

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

and

T_2 = 87 + 273 = 360\text{ K}

Gas laws require thermodynamic (absolute) temperature in Kelvin.

State the combined gas law relating initial and final states.

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

The mass of gas is fixed while pressure, volume, and temperature all change.

Rearrange the equation to solve for the final volume

V_2

V_2 = V_1 \times \left(\frac{P_1}{P_2}\right) \times \left(\frac{T_2}{T_1}\right)

Isolating the target unknown variable.

Substitute the given numerical values into the rearranged equation.

V_2 = 0.25\text{ m}^3 \times \left(\frac{1.6 \times 10^5\text{ Pa}}{8.0 \times 10^4\text{ Pa}}\right) \times \left(\frac{360\text{ K}}{300\text{ K}}\right) = 0.25 \times 2.0 \times 1.2 = 0.60\text{ m}^3

Simplifying the ratios yields the correct final volume.

Key Concept

Combined Gas Law

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