Answer:
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
