By "isotropic", you presumably mean "Isentropic", meaning no heat transfer across the boundary.
The perfect gas equation you quoted is actually for Isothermal, meaning no temperature change. Let me demonstrate:
(P1*V1/T1) = (P2*V2/T2); the full perfect gas equation.
If temperature is constant (i.e. Isothermal), then T1 = T2 so both T1 and T2 can be removed from the equation to leave:
P1*V1 = P2*V2 (which is what you've got).
As long as the changes are happening slowly, leaving plenty of time for heat transfer (which I would suggest is the case), then you have Isothermal processes, in which case you can indeed use your equation.
Whoops! That's what I meant: Isothermal. If the pressure changes a few psi over the course of a second or two, would that be sufficient enough to assume isothermal activity? I'd assume so, myself, but I want to be sure. Note that this change would be no more than about 15% of the initial pressure at any time.
Say you have a vessel containing 10 litres of air at a pressure of 2 bar gauge, and its pressure increases by about 15% over a couple of seconds. Now I would say that the process is almost certainly not isothermal but, on the other hand, neither is it adiabatic. I think I'll just sit on the fence for that one! However, in the real world, it makes very little difference (i.e. no difference) to me as a practising engineer since the difference is so little.
To be more general, instead of using P1*V1 = P2*V2 (as for your isothermal case), lets assume the correct relationship is P1*(V1^n) = P2*(V2^n). If the process is isothermal, then n = 1; if it is adiabatic (a more practical / likely version of the very ideal isentropic), then n = 1.4.
Do the sums yourself; they should agree with mine thus:
Isothermal case: pressure increases by about 15%, volume decreases to about 87%.
Adiabatic case: pressure increases by about 15%, volume decreases to about 90% (the volume has not decreased so much because no heat transfer means the air's temperature has increased under compression, so it exerts a higher pressure at each volume than it otherwise would have done).
For my money, there's very little difference; what do you think?
I suppose I'd have to determine exactly how critical estimating the volume will be. My guess is that it's not extremely vital in my case. Running my simulation for either case, I haven't noticed very much difference between the two cases, so that seems more or less in line with those results.
For a simulated isothermal condition to take place, the change in pressure of a few psi in the course of one or two second should not be considered isothermal because heat transfer would not have had time to keep gas at constant temperature; in thirty minutes I would say yes.
Can you help with a variation of the same problem? I recently have been working with an air charged piston type accumulator. I am trying to understand how volume/pressure relationship is affected by differing air qualities on the air side; for instance, in the ideal case where the air charge is instrument quality and dry (-40 degree dew point), how will accumulator performance degrade if air is saturated with water vapor.
I assume my conditions to be adiabatic.
Thanks in advance for your advice.
Tom
