• ewang
In summary: This is the opposite of the ## dU = \delta Q + p dV ## that is usually written in engineering. (I am an engineer, but I come from a physics family.)In summary, for an adiabatic process starting at V = 0.024 m^3 and 101325 Pa and ending at 0.0082 m^3 and 607950 Pa, the work done by the gas is negative. The work can be calculated using the equation pV * (Vf(1 - gamma) - Vi(1 - gamma))/(1-gamma). When plugging in the numbers, the result is -3823.6 J. This is due to the standard thermodynamics convention of

#### ewang

Homework Statement
Find the work done by the gas (or on the gas) for an adiabatic process starting at V = 0.024 m^3 and 101325 Pa and ending at 0.0082 m^3 and 607950 Pa. The working gas is helium
Relevant Equations
Work for adiabatic = area under pV diagram
p1V1^gamma = p2V2^gamma
This is a relatively simple problem, but I'm not getting the right answer. For adiabatic compression, work on gas is positive, since work on gas = ΔEth and the adiabatic process moves from a lower isotherm to a higher one. Integrating for work gives:
pV * (Vf(1 - gamma) - Vi(1 - gamma))/(1-gamma)
I believe this is correct, but when I plug in the numbers, I'm getting a negative number:
101325 Pa * 0.024 m3 * ((0.0082 m3)1 - 1.67 - (0.024 m3)1 - 1.67)/(1 - 1.67)
= -3823.6 J

ewang said:
Homework Statement:: Find the work done by the gas (or on the gas) for an adiabatic process starting at V = 0.024 m^3 and 101325 Pa and ending at 0.0082 m^3 and 607950 Pa. The working gas is helium
Relevant Equations:: Work for adiabatic = area under pV diagram
p1V1^gamma = p2V2^gamma

This is a relatively simple problem, but I'm not getting the right answer. For adiabatic compression, work on gas is positive, since work on gas = ΔEth and the adiabatic process moves from a lower isotherm to a higher one. Integrating for work gives:
pV * (Vf(1 - gamma) - Vi(1 - gamma))/(1-gamma)
I believe this is correct, but when I plug in the numbers, I'm getting a negative number:
101325 Pa * 0.024 m3 * ((0.0082 m3)1 - 1.67 - (0.024 m3)1 - 1.67)/(1 - 1.67)
= -3823.6 J

Nevermind, work is negative integral oops. I was staring at this for the longest time.

The standard thermodynamics convention of signs is the Clausius convention
ΔU = Q - W
the variation of internal energy = Heat added to the system - Work done

Thus when the gas expands we have positive work

Lnewqban
ewang said:
Homework Statement:: Find the work done by the gas (or on the gas) for an adiabatic process starting at V = 0.024 m^3 and 101325 Pa and ending at 0.0082 m^3 and 607950 Pa. The working gas is helium
Relevant Equations:: Work for adiabatic = area under pV diagram
p1V1^gamma = p2V2^gamma

For adiabatic compression, work on gas is positive
Right. work done BY gas is negative. The 1st law is usually written ## dU = \delta Q - p dV ## in physics.