# Solving Thermodynamics Doubt - João

• jaumzaum
In summary: OUNT OF WORK DONEIn summary, the question asks for the amount of work done when one mole of an ideal gas expands isothermally from 1atm to 0.1 atm against a constant external pressure of 0.1 atm at 298K. The first answer is correct, taking into account the constant external pressure, while the second answer is incorrect because it uses the internal gas pressure instead of the external pressure. The definition of work is the area under the PV diagram, and in this case, the thermodynamic work is done by the gas against the external pressure.

#### jaumzaum

I was solving the question below :

"A cylinder have a piston whose mass is insignificant and moves without friction. One mole of an ideal gas is confined into the cylinder. If we expand isotermally the gass at temperature 298K, against a constant external pressure of 0.1 atm, the gas pressure goes from 1atm to 0.1 atm. Calculate the work need for the expansion."

I have 2 answers, and I don't know why the second one is wrong:
1) Pressure is constant. W = P(ext)(V2-V1) = P(ext) nRT(1/P2-1/P1) = 0,9.1.8,31.298 = 2.23kJ

2) Gas Pressure is P, so the difference of pressure is P1=(P-0,1).

W = ∫P1.dv.
V = nRT/P
dv = -nRT/P²dP
W = ∫-(P-0.1)nRT/P² dP = 2476(ln(P1/P2) - 0.1(P2-P1)) = 2476(2.3+0.09) = 5,91kJ

Why the second is wrong, and more important, why the first one is right? What is the definition of work?

[]'s
João

The Work is the area under the PV diagram.

What makes you think (1) is correct and (2) is not?

jaumzaum said:
I was solving the question below :

"A cylinder have a piston whose mass is insignificant and moves without friction. One mole of an ideal gas is confined into the cylinder. If we expand isotermally the gass at temperature 298K, against a constant external pressure of 0.1 atm, the gas pressure goes from 1atm to 0.1 atm. Calculate the work need for the expansion."

I have 2 answers, and I don't know why the second one is wrong:
1) Pressure is constant. W = P(ext)(V2-V1) = P(ext) nRT(1/P2-1/P1) = 0,9.1.8,31.298 = 2.23kJ

2) Gas Pressure is P, so the difference of pressure is P1=(P-0,1).

W = ∫P1.dv.
V = nRT/P
dv = -nRT/P²dP
W = ∫-(P-0.1)nRT/P² dP = 2476(ln(P1/P2) - 0.1(P2-P1)) = 2476(2.3+0.09) = 5,91kJ

Why the second is wrong, and more important, why the first one is right? What is the definition of work?

[]'s
João
In 2. you are using the internal pressure of the gas rather than the external pressure. The thermodynamic work here is the work done by the gas on the external environment (ie. against the external pressure). In effect, the excess internal pressure is not used to do mechanical work.

This is a non-reversible expansion because there is a significant difference between the internal gas pressure and the external pressure. This excess pressure causes a very rapid initial outward expansion but does not add to the total work that is done. That work is simply W = Pext(ΔV).

AM