A rigid gas vessel sealed on a meteorological satellite initially contains a fixed volume … — JAMB UTME

A rigid gas vessel sealed on a meteorological satellite initially contains a fixed volume of an ideal gas at a pressure of $1.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . During an orbital maneuver, solar radiation heats the vessel to $327^\circ\text{C}$ , while a small vent valve simultaneously allows $20\%$ of the gas molecules to escape. Assuming the volume of the vessel remains constant, what is the final pressure of the gas inside the vessel?

$2.40 \times 10^5\text{ Pa}$ Answer
B
$1.45 \times 10^6\text{ Pa}$
C
$3.00 \times 10^5\text{ Pa}$
D
$6.00 \times 10^4\text{ Pa}$

Answer

2.40 \times 10^5\text{ Pa}

According to the ideal gas law (

PV = nRT

), when volume is held constant, pressure is directly proportional to both the number of moles of gas present and the absolute temperature in Kelvin (

P \propto nT

). First, convert the initial and final temperatures to Kelvin:

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

and

T_2 = 327 + 273 = 600\text{ K}

. Because

20\%

of the gas escapes,

80\%

remains inside the vessel, so

n_2 = 0.80n_1

. Calculating final pressure yields

P_2 = P_1 \times (n_2/n_1) \times (T_2/T_1) = 1.50 \times 10^5 \times 0.80 \times (600/300) = 2.40 \times 10^5\text{ Pa}

Step-by-Step Solution

Convert all given temperatures from degrees Celsius to absolute temperature in Kelvin.

T_1 = 27 + 273.15 = 300\text{ K}

T_2 = 327 + 273.15 = 600\text{ K}

Gas laws and the ideal gas equation require absolute temperature in Kelvin.

Determine the fraction of gas remaining in the rigid vessel.

20\%

escapes, remaining fraction

n_2 = (1 - 0.20)n_1 = 0.80n_1

Pressure depends on the quantity of gas remaining inside the fixed volume.

Apply the ideal gas equation

PV = nRT

for constant volume

V

P_2 = P_1 \times \left(\frac{n_2}{n_1}\right) \times \left(\frac{T_2}{T_1}\right)

Since volume is constant, pressure is directly proportional to the product of amount of gas and absolute temperature.

Substitute known values to compute the final pressure

P_2

P_2 = (1.50 \times 10^5\text{ Pa}) \times 0.80 \times \left(\frac{600\text{ K}}{300\text{ K}}\right) = 2.40 \times 10^5\text{ Pa}

Multiplying the initial pressure by the mole fraction ratio and temperature ratio gives the final gas pressure.

Key Concept

Ideal Gas Equation and Combined Gas Behavior

Estimated Time:2m 0s

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