A rigid container A of volume 0.060\text{ m}^3 contains an ideal gas at an absolute pressu… — JAMB UTME

A rigid container $A$ of volume $0.060\text{ m}^3$ contains an ideal gas at an absolute pressure of $4.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . Container $A$ is connected via a narrow tube of negligible volume with a valve to an evacuated rigid container $B$ of volume $0.040\text{ m}^3$ . The valve is opened, allowing gas to flow between the containers until equilibrium is established. If container $A$ is maintained at $27^\circ\text{C}$ while container $B$ is maintained at $127^\circ\text{C}$ , what is the final equilibrium pressure of the gas in kilopascals (kPa)?

Answer: 300 kPa

Answer

300 kPa

Converting both temperatures to Kelvin (

300\text{ K}

and

400\text{ K}

) and applying mole conservation (

n_{\text{total}} = n_A + n_B

) yields a final uniform pressure of

3.00 \times 10^5\text{ Pa}

, which equals

300\text{ kPa}

Step-by-Step Solution

Convert temperatures from degrees Celsius to kelvins.

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

and

T_B = 127^\circ\text{C} + 273 = 400\text{ K}

Gas laws and calculations using the ideal gas equation require absolute temperatures in Kelvin.

Calculate the initial number of moles of gas present in the system.

n_{\text{total}} = \frac{p_A V_A}{R T_A} = \frac{(4.50 \times 10^5\text{ Pa})(0.060\text{ m}^3)}{R(300\text{ K})} = \frac{90}{R}

Container B is initially evacuated, meaning all gas molecules originate from container A.

Express the total number of moles at final equilibrium in terms of final pressure

P

n_{\text{final}} = n_A + n_B = \frac{P V_A}{R T_A} + \frac{P V_B}{R T_B} = \frac{P}{R} \left(\frac{0.060}{300} + \frac{0.040}{400}\right) = \frac{P}{R} (2.0 \times 10^{-4} + 1.0 \times 10^{-4}) = \frac{3.0 \times 10^{-4} P}{R}

At equilibrium, the pressure

P

becomes uniform throughout both interconnected containers.

Equate the initial and final total moles to solve for the equilibrium pressure

P

\frac{90}{R} = \frac{3.0 \times 10^{-4} P}{R} \implies P = \frac{90}{3.0 \times 10^{-4}} = 3.00 \times 10^5\text{ Pa} = 300\text{ kPa}

The total mass and number of moles of gas are conserved within the sealed system.

Key Concept

Conservation of total moles in interconnected gas containers governed by the ideal gas equation

PV = nRT

Estimated Time:2m 30s

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