# If a gas begins at "600 cm"^3 and its pressure increases by "30 kPa" from "100 kPa" of pressure in one minute, what is the average rate of change in volume per unit time?

May 24, 2017

At constant temperature and mols of ideal gas...

...since we know (DeltaP)/(Deltat, we must assume the rates of change are linear (which is not necessarily true!!) in an interval of $\text{1 min}$ time. Then, since volume and pressure are state functions (just like enthalpy, entropy, gibbs' free energy, etc),

$\Delta \left(P V\right) = {\cancel{\Delta \left(n R T\right)}}^{\text{assumed 0}} = {V}_{1} \Delta P + {P}_{1} \Delta V + \Delta P \Delta V$

gives the change in both pressure and volume at the same time.

Since $P V = \text{const}$, the net change $\Delta \left(P V\right)$ is zero, i.e. ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$, Boyle's law.

Dividing by the change in time, we get:

$\frac{\Delta \left(P V\right)}{\Delta t} = 0 = {P}_{1} \frac{\Delta V}{\Delta t} + {V}_{1} \frac{\Delta P}{\Delta t} + \frac{\Delta V}{\Delta t} \cdot \Delta P$

$= \left({P}_{1} + \Delta P\right) \frac{\Delta V}{\Delta t} + {V}_{1} \frac{\Delta P}{\Delta t}$

Therefore, in one minutes' time, the second pressure became $\text{130 kPa}$, since it increased by $\text{30 kPa}$ from ${P}_{1} = \text{100 kPa}$:

$\textcolor{b l u e}{\frac{\Delta V}{\Delta t}} = - \frac{{V}_{1} \left(\frac{\Delta P}{\Delta t}\right)}{{P}_{1} + \Delta P}$

$= - \left(\text{600 cm"^3 ("30 kPa"/"min"))/("100 kPa" + "30 kPa}\right)$

$\approx \textcolor{b l u e}{- \text{138.5 cm"^3/"min}}$

at constant temperature and mols of gas.

This should make physical sense. In one minutes' time, if the pressure becomes $1.3$ times what it was, then the volume drops so that it becomes $\frac{1}{1.3}$ times what it was, which keeps $P V$ constant.

What is the volume after this $\text{1 min}$ interval, then?