Question

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?

Answer 1

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

Step-by-step 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