Physics Thermodynamics - Finding Work

[misa]

Homework Statement

A 1.20 mol sample of an ideal diatomic gas at a pressure of 1.20 atm (P₁) and temperature of 380 K (T₁) undergoes a process in which its pressure increases linearly with temperature. The final temperature and pressure are 680 K (T₂) and 1.83 atm (P₂).

(b) Determine the work done by the gas.

Homework Equations

(T₂ - T₁) / (P₂ - P₁) = constant

V₂ / V₁ = P₁T₂ / P₂T₁

PV = nRT

W = integral of PV

The Attempt at a Solution

The answer is supposed to be [nR(P₁T₂ - P₂T₁) ln(P₂/P₁)] / (P₂ - P₁)
but I can't figure out how to get there despite the hours I have poured into this problem. All I know is that volume is not constant in order for there to be work done. Also, I have the vague idea of finding the linear equation or relationship between T and P using that equation to plug in T in PV = nRT. Then, technically, there would be a graph for PV and I could solve for V₁ and V₂, which can be the limits of integration for the integral of PV (to get work). But isn't that way too complicated, especially as an integral...so I was wondering whether someone else has any idea.

Please help explain, and thank you!

 

[Delphi51]

I certainly understand your frustration!
If P varies linearly with T, you would expect P = kT. Put that in V = nRT/P and you get that V is a constant, which is wrong. Looks like the question is inconsistent!

The other meaning of "linear" would be P = kT + b so V = nRT/(kT+b) which doesn't seem likely. It does give quite a complex answer.
Since P and V both vary, dW = PdV + VdP
I worked the integral quickly and got
W = -nRb(P2 - P1) + nR/k*[P2-P1 -ln(P2/P1)]
Could well be errors in that! It doesn't look much like the given answer, does it?

 

[misa]

Ah, thank you so much for the insight. (I'll have to try and work that out when I can find the time.)