Finding Pressure with constant temperature and varying volumeby joe465 Tags: constant, pressure, temperature, varying, volume 

#1
Jun1912, 05:10 PM

P: 94

1. The problem statement, all variables and given/known data
A 3litre vessel contains gas at a pressure of 200 kPa. Five litres of gas at the same pressure is forced into the vessel. Calculate the new pressure of the gas. 2. Relevant equations P1V1 = P2V2 3. The attempt at a solution P1V1 = P2V2 (Keeping Temp = Constant) V2 = 3 + 5 = 8 litres 3 * 200 = P2 * 8 P2 = 600/8 = 75 kPa I have been told this answer is wrong but unsure why if anyone could help me out, would greatly appreciate it. Thanks, Joe 



#2
Jun1912, 06:18 PM

HW Helper
PF Gold
P: 1,842





#3
Jun2012, 05:04 AM

P: 94

Thanks for your reply, well the question has been copied word for word and just says gas so i presume it would be right to assume its an ideal gas.
So you suggest V1 would be 8L and the 5L is forced into the 3L section making V2 3L therfore making the pressure rise. P1V1 = P2V2 (Keeping Temp = Constant) V2 = 8  5 = 3 litres 8 * 200 = P2 * 3 P2 = 1600/3 = 533.33 kPa That looks like a more realistic outcome, what do you think? 



#4
Jun2012, 04:27 PM

HW Helper
PF Gold
P: 1,842

Finding Pressure with constant temperature and varying volume 



#5
Jun2012, 04:30 PM

P: 94

Thankyou very much, it literally only states what it does in the question and doesnt suggest temperature change.
Would the question emply or give a temperature if this was required? Thanks, Joe 



#6
Jun2012, 05:04 PM

P: 428

PV=nRT.
Process 1, adding gas at fixed pressure, n changes, V changes, T stays same. Process 2, with new V, we compress it to old V. Thus in PV=nRT, the pressure and temp are ambiguous. We must compress it in a certain way. A temp bath would keep the temp constant. A thermos would stop heat exchange (not the same as temp bath). A free piston on top would keep pressure the same. The problem is ambiguous since it doesn't tell us which process to use for process 2. That is, either isothermic, adiabatic or isobaric. 



#7
Jun2012, 05:08 PM

HW Helper
PF Gold
P: 1,842

Even if it doesn't state anything about temperature, then I'm still guessing that your answer is fine, and the process is assumed to be isothermal. On the other hand, if the process is adiabatic, then things get a little more complicated. But I doubt this is the case, since the problem statement never mentioned whether the gas was monatomic, diatomic, or whatnot. If your coursework hasn't discussed this stuff (yet), I'd assume that the temperature stays constant, and your answer is correct. 


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