Problem 16.48

16.48. a)
We need to come up with a relation between the initial and final pressures and volumes.
Summing the forces initially, when the gas is at $250{}^{\circ}C$, we find
(P_f-P_i)A - kh = 0
or $P_{f} = P_{i} + \frac{kh}{A}$
We can see from inspection that the volumes are related by
V_f = V_i + Ah
Now, we can use the relation
$\frac{P_{i}V_{i}}{T_{i}} ~=~ \frac{P_{f}V_{f}}{T_{f}}$
Plugging in our expressions for the initial pressure and volumes,
$\frac{(P_{i}+\frac{kh}{A})(V_{i}+Ah)}{T_{f}} ~=~ \frac{P_{i}V_{i}}{T_{i}}$
Plugging in numbers, we find
$\frac{\left((101.3 kPa)+(2\times 10^{5})h\right)(5\times 10^{-3}+.01 h)}{523} ~=~ \frac{(101.3kPa) (5\times 10^{-3})}{293}$
This reduces to a quadratic equation,
2000h + 2013h - 397 = 0
or h = .169 m

b)
The final pressure is given by
$P_{f} = P_{i} + \frac{kh}{A}$
$P_{f} = 101.3 kPa + \frac{(2000)(.169)}{.01}$
P_f = 135 kPa