Air is pumped into the tubes of a cycle rickshaw at a pressure of `2atm`. The volume of each tube at this pressure is `0.002m^(3)`. One of the tubes gets punctured and the volume of the tube reduces to `0.0005m^(3)`.How many moles of air have leaked out? Assume that the temperature remains constant at 300K and that the air behaves as an ideal gas.

Explanation

`P_1 = 2 atm = 2 xx (10^5)Pa`
` V_1 = 0.002(m^3), T_1 = 300K`
` (P_1)(V_1) = (n_1)(RT_1)`
` rArr (n_1) = ((P_1)(V_1)/ (RT_1)) = (2 xx (10^5)xx 0.002/ 8.3 300)`
` = (2 xx (10^5) xx 2 xx (10^-3)/ 8.3 xx (10^2) xx 3)`
` = (4/ 8.3 xx 3) = 0.1606.`
` Again, (P_2) = 1 atm = 10^5Pa`
` V_2 = 0.0005 (m^3), (T_2)= 300K`
` ((P_2)(V_2))= (n_2)(RT_2)`
` rArr (n_2) = ((P_2)(V_2)/(RT_2)) = ((10^5) xx 0.0005 / 8.3 xx 300)`
` = ((10^5) xx 5 xx (10^-4)/ 8.3 xx 3 xx (10^2))`
` = (5/ 3 xx 83) = 0.02`
` Delta m = moles leaked out `
` = 0.16 - 0.02 = 0.14` .