Problem 16.49

16.49. a)
In equillibrium, the forces are equal.
mg + P_aA = P_gasA
Where P_a is atmospheric pressure. The pressure of the gas is given by
$P_{gas} = \frac{nRT}{V}$
Where we used the volume of the gas was hA.
$P_{gas} = \frac{nRT}{hA}$
so, then in terms of the height,
$\frac{nRT}{hA} = \frac{mg}{A} + P_{a}$
or $h = \frac{nRT}{(mg + P_{a}A)}$

b)
Plugging in the numbers, we find
$h = \frac{(.200 mol)(8.31)(400{}^{\circ}K)}{(20)(9.8) + (1.0 \times 10^{5})(.008)}$
h = .662 m