Question about work done by expanding pressurized gas

AI Thread Summary
The discussion centers on the work done by expanding pressurized gas, specifically addressing the integral of 1/V*dV, which suggests infinite work as volume approaches infinity. Participants clarify that while the equation indicates potential infinite work, it is constrained by thermodynamic laws, particularly the first law of thermodynamics. The heat flow into the system must balance the work done and the change in internal energy, indicating that infinite work is only possible with unlimited heat supply. Additionally, a new member seeks guidance on calculating the work ratios for air in a piston cylinder during compression and expansion, given limited initial data. The conversation highlights the importance of understanding thermodynamic principles in practical applications.
JerryG
Messages
58
Reaction score
0
In my thermodynamics class, we were shown an equation that gives the work done by expanding pressurized gas given as the integral from v1 to v2 of C/V*dV where C is a constant and v1 and v2 are initial and final volumes respectively. My question has to do with the fact that this is basically the integral of 1/V*dV which implies that as a gas expands to infinity, it will do infinite work which then implies that pressurized gas has infinite potential energy which I know cannot be true. Can anyone explain what is wrong with this logic? I asked the professor, but she just said it had something to do with thermodynamic laws not applying to stuff that goes to infinity.
 
Physics news on Phys.org
Hi, JerryG

JerryG said:
In my thermodynamics class, we were shown an equation that gives the work done by expanding pressurized gas given as the integral from v1 to v2 of C/V*dV where C is a constant and v1 and v2 are initial and final volumes respectively.

Equation of state for ideal gas PV=NkT stands for your case. P=NkT / V = C /V
As C is a constant, T is constant so the system is in heat bath of constant temperature T. Infinite energy is supplied from heat bath to gas during its infinite expansion.

Regards
 
Last edited:
Further to what sweet springs has said: apply the first law:

\Delta Q = \Delta U + W = \Delta U + C\ln\frac{V_2}{V_1}

The heat flow into the system less the change in internal energy has to equal the work done by the system. There is a limit to the internal energy change. So there can be no limit to the work done only if there is no limit to the heat flow into the system.

AM
 
Hi I am a new member and this is my first post. I am currently taking an engineering Thermodynamics class and am having some trouble with an assigned problem dealing with air in a piston cylinder.

Process 1-2: Air is compressed PV^n=constant
Process 2-3: Air expands P=constant until V3=V1

given r=5=V1/V2 and T1=300K

determine ratio of the work of expansion to the work of compression in terms of r and n when
n does not equal 1 and when n=1 and evaluate when A)n=1.4 and B)n=1

I had no problem finding Wexp/Wcomp when n does not equal one
(r-1)(n-1) / (1-r^(1-n)) A)3.37

and Wexp/Wcomp when n=1 (r-1)/ln[r] B)2.48

part 2 check answers by finding the actual Wexpansion and Wcompression and forming the the ratios for each case.

my question is given only T1 and V1/V2=5 do I have enough information to actually solve for anything other then the ratio itself in each case?
Ideal gas so PV=mRT and Specific heat equations are what I have been working from and I have a Property Table book which supplies u1 and h1 values.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Back
Top