Why Does Adiabatic Compression Yield a Negative Work Calculation?

[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 * (V_f^{(1 - gamma)} - V_i^{(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]

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 * (V_f^{(1 - gamma)} - V_i^{(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.

 

[valentin bogatu]

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

 

[rude man]

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.