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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
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Solution
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
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