# 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?

##### 1 Answer

At constant temperature and mols of ideal gas...

...since we know

#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

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

#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

What is the volume after this