Question

The pressure P, Volume V and Temperature T of a mole of an ideal gas are related by the
equation PV = 8.31 T. Find the rate at which the pressure is changing when the
temperature is 300k and increasing at a rate of 0.1 per second and the volume is 100 L and
increasing at the rate of 0.2 per second.

Answer 1

P=8.31 T/V; T = 300K; V = 100L = 0.1 m^3;

$$\frac{d T}{d t} = 0.1 K/sec; \frac{d V}{d t} = 0.2 L/sec = 2*10^{-4} m^3/sec$$

$$\frac{d P}{d t} = \frac{\partial P}{\partial T} \frac{d T}{d t} + \frac{\partial P}{\partial V} \frac{d V}{d t} = \frac{8.31}{V}\frac{d T}{d t} -\frac{8.31T}{V^2}\frac{d V}{d t}$$

$$\frac{d P}{d t} =\frac{8.31}{0.1} 0.1 - \frac{8.31*300}{0.1*0.1} 2*10^{-4}=-41.55 Pa/sec$$

The pressure decreases at the rate 41.55 Pa per second.