Question

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?

Answer 1

At constant temperature and mols of ideal gas...

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

`Delta(PV) = cancel(Delta(nRT))^("assumed 0") = V_1DeltaP + P_1DeltaV + DeltaPDeltaV`

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

Since `PV = "const"`, the net change `Delta(PV)` is zero, i.e. `P_1V_1 = P_2V_2`, Boyle's law.

Dividing by the change in time, we get:

`(Delta(PV))/(Deltat) = 0 = P_1(DeltaV)/(Deltat) + V_1(DeltaP)/(Deltat) + (DeltaV)/(Deltat)cdot DeltaP`

`= (P_1 + DeltaP)(DeltaV)/(Deltat) + V_1(DeltaP)/(Deltat)`

Therefore, in one minutes' time, the second pressure became `"130 kPa"`, since it increased by `"30 kPa"` from `P_1 = "100 kPa"`:

`color(blue)((DeltaV)/(Deltat)) = -(V_1((DeltaP)/(Deltat)))/(P_1 + DeltaP)`

`= -("600 cm"^3 ("30 kPa"/"min"))/("100 kPa" + "30 kPa")`

`~~ color(blue)(-"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 `1/1.3` times what it was, which keeps `PV` constant.

What is the volume after this `"1 min"` interval, then?