gas in a piston-cylinder assembly with initial volume 100cm^3 is compressed from 1 atm to 2atm. If the process happens at constant temperature, what is the work involved in the process?
i think there is no work involved since T is constant. but i'm not sure. please help!
2007-10-10 07:21:05 · 6 answers · asked by Effrem 1 in Science & Mathematics ➔ Engineering
i know but if i don't have two volumes given and pressure isn't constant can i use it?
2007-10-10 07:29:14 · update #1
can i treat the gas as an ideal gas
2007-10-10 07:31:26 · update #2
Work is the integral of P dV for a constant-temperature expansion or compression.
EDIT: If pressure is not kept constant during the compression, then you need to evaluate the integral where pressure is a function of volume at constant pressure according to the Ideal Gas Law. You can safely assume Ideal Gas Behavior for low pressures.
To save you some time, the integral for work done (assuming Ideal Gas) is:
W = n R T ln (V2 / V1) = n R T ln (P1 / P2)
where n = number of moles, V1 = starting volume, V2 = ending volume, P1 = starting pressure, P2 = ending pressureR = ideal gas constant, T = temperature
Ideal gas law: PV = n RT
2007-10-10 07:24:26 · answer #1 · answered by John 7 · 2⤊ 0⤋
Start with the definition of work: W=F*s.
When you compress the gas in the cylinder, how much force F does it take (as a function of pressure)? By how much (s) do you have to move the cylinder to double the pressure?
What is dW(p)=F(p)*ds, the work you have to perform to compress the gas a little bit along the way?
If you integrate that while you are going from the initial pressure to the final pressure, what do you get?
Good luck!
2007-10-10 14:28:14 · answer #2 · answered by Anonymous · 0⤊ 1⤋
Yes there is work involved. If the chamber is connected to a
reservoir at a constant temperature and you let the piston
expand, it will and so work will be done. If it is connected
to a reservoir and you compress the piston, you are doing work on the system.. Think PV=nRT. This is PV work.
W=-/int(p1,p2)PdV, P=~1/vP1V1=P2V2, V2=50
W = -nRT [ln(P1÷P2)] =-nRTln.5
2007-10-10 14:53:19 · answer #3 · answered by jim m 5 · 0⤊ 0⤋
If P*V = Constant , then the external work done = W = 144*P1*V1*(InV2/V1) foot-pounds.
This is derived from W = 144*Integral from V1 to V2 of the function P*dv.
If P = Constant, then the external work done = W = 144*P*(V2-V1)
If V = Constant, then the external work done = W = 0
2007-10-10 22:02:18 · answer #4 · answered by gatorbait 7 · 0⤊ 0⤋
The approaches you already have will work out the problem, my guess is, that the 'constant temperature' reference is meant to state that no 'external temperature' affects the process.
2007-10-10 17:22:47 · answer #5 · answered by mavis b 4 · 0⤊ 0⤋
Wkn=(P1)(V1)[ ln(P1/P2) ]
2014-05-16 05:14:51 · answer #6 · answered by intut 1 · 0⤊ 0⤋