Question

Air is pumped into the tubes of a cycle rickshaw at a pressure of 2 atm. The volume of each tube at this pressure is 0.002 m³. One of the tubes gets punctured and the volume of the tube reduces to 0.0005 m³. How many moles of air have leaked out? Assume that the temperature remains constant at 300 K and that the air behaves as an ideal gas.

Use R = 8.3 J K^-1 mol^-1

Answer 1

Here,
P₁ = 2 × 10⁵ pa
V₁ = 0.002 m³
V₂ = 0.0005 m³
T₁ = T₂ = 300 K
Number of moles initially , n₁ = \[\frac{P_1 V_1}{R T_1} \]
⇒  n₁ = \[\frac{2 × {10}^5 \times 0.002}{8.3 × 300} \]
⇒ n₁ = 0.16
Applying equation of state, we get
P₂ V₂ = n₂ RT
Assuming the final pressure becomes equal to the atmospheric pressure, we get
P₂ = 1.0 × 10⁵ pa
⇒ n₂ = \[\frac{P_2 V_2}{RT} \]
⇒ n₂ = \[\frac{1.0 × {10}^5 × 0.0005}{8.3 × 300} \]
⇒ n₂ = 0.02
Number of leaked moles= n₂ - n₁
                                      = 0.16 -0.02 
                                      = 0.14