Consider a day in the physics lab room with room temperature of 20 0C and pressure of 1.0 atm. A container of 3.5 L is left open to the air for a long time. The container is then sealed and place on a Bunsen burner until its temperature reaches 95 0C. It is then open. How many moles of air escape?

The number of moles of air that escape is approximately 0.027 moles

Explanation:

The room temperature, T₁ = 20 °C = 298.15 K

Atmospheric pressure, P₁ = 1.0 atm

The volume of the container, V₁ = 3.5 L

The final temperature of the air in the container after heating on the Bunsen burner, T₂ = 95 °C = 368.15 K

The container opened finally

The ideal gas equation is P·V = n·R·T

∴ n = P·V/(R·T)

Where;

R = The universal gas constant = 0.08205 L·atm/(mol·K)

Therefore, we get;

n₁ = 1.0 × 3.5/(0.08205 × 298.15) ≈ 0.143

The number of moles, n₁ ≈ 0.143 moles

When the gas is heated to 95 °C, the number of moles becomes

n₂ =  P₂·V₂/(R·T₂)

P₂ = 1.0 atm atmospheric pressure, V₂ = 3.5 L, the volume of the container, T₂ =

∴ n₂ = 1.0 × 3.5/(0.08205 × 368.15) ≈ 0.116

The number of moles of air remaining in the container, n₂ ≈ 0.116 moles

The number of moles of air that escape, n = n₁ - n₂

∴ n = 0.143 - 0.116 = 0.027

The number of moles of air that escape, n ≈ 0.027 moles

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