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InterviewSolution
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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. |
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Answer» 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 |
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