A vessel of volume `8.0 xx 10^(-3) m^(3)` contains an ideal gas at `300 K` and `210 kPa`. The gas is allowed to leak till the pressure falls to `135 kPa`. Calculate the amount of the gas (in moles) leaked assuming the temperature to remain constant.

Explanation

There is no change in volume and temperature of the gas during leakage. However, the number of moles of the gas or its mass changes.
According to ideal gas equation, `PV = nRT` or `n = (PV)/(RT)`
Before leakage, `n_(1) = (P_(1)V)/(RT)`
After leakage, `n_(2) = (P_(2)V)/(RT)`
`:.` Change in no. of moles i.e., `(n_(1) - n_(2)) = (V(P_(1) - P_(2)))/(RT)`
According to available data,
`V = 8.0 xx 10^(-3)m^(3) , P_(1) = 210 xx 10^(3) Nm^(-2)`,
`P_(2) = 135 xx 10^(3) Nm^(-2) , R = 8.314 Nm K^(-1) mol^(-1)`,
`T = 300 K`
Substituting the values, `Deltan = ((8.0 xx 10^(-3)m^(3)) xx (210 - 135) xx (10^(3) Nm^(-2)))/((8.314 NmK^(-1)mol^(-1)) xx 300K) = 0.240 mol`.
`:.` No. of moles of gas leaked `= 0.240` mol