InterviewSolution
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InterviewSolution
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Calculate the pressure exerted by one mole of `CO_(2)` gas at `273 K` van der Waals constant `a=3.592 dm^(6) atm mol^(-2)`. Assume that the volume occupied by `CO_(2)` molecules is negligible. |
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Answer» The van der Waals equation for one mole of a gas is `(P+(a)/(V^(2)))(V-b)=RT` It is given that the volume occupied by `CO_(2)` molecules is negligible. Hence, the equation of state becomes `(P+(a)/(V^(2)))(V)=RT` or `p=(RT)/(V)-(a)/(V^(2))` Assuming `V_(m)=22.414 dm^(3) mol^(-1)`, we get `p=((8.314 kPa dm^(3) K^(-1) mol^(-1))(273 K))/((22.414 dm^(3) mol^(-1))-(3.592 dm^(6)atm mol^(-2))/((22.414 dm^(3) mol^(-1))^(2))` `=101.246 kPa-7.15xx10^(-3)atm` `=101.246 kPa-(7.15xx10^(-3)atm)((101.325 kPa)/(1 atm))` `=101.264 kPa-0.724 kPa` `=100.601 kPa` |
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