A straight line thermodynamic transformation

[T-7]

Hi,

Homework Statement

One mole of a diatomic ideal gas performs a transformation from an initial state for which the temp. and vol. are respectively 291K and 21000 cc, to a final state in which temp. and vol. are 305K and 12700cc - a straight line in the P,V diagram. Find work performed, and heat absorbed by sys.

The Attempt at a Solution

I'm wondering which is the best approach. I could calc. the pressures and use them to express the pressure as a linear function of the volume, and integrate that between the two volumes (int(P dV)) to get the work.

But perhaps I'm just not thinking, and there is a more obvious, elegant approach to this whole question (?)

Cheers.

 

[HungryChemist]

Hi,

Homework Statement

One mole of a diatomic ideal gas performs a transformation from an initial state for which the temp. and vol. are respectively 291K and 21000 cc, to a final state in which temp. and vol. are 305K and 12700cc - a straight line in the P,V diagram. Find work performed, and heat absorbed by sys.

The Attempt at a Solution

I'm wondering which is the best approach. I could calc. the pressures and use them to express the pressure as a linear function of the volume, and integrate that between the two volumes (int(P dV)) to get the work.

But perhaps I'm just not thinking, and there is a more obvious, elegant approach to this whole question (?)

Cheers.

If you know the process is done in straight line in P-V diagram, to get the work done, just calculate the area under that straight line path. Can't get any easier. But anyway, that's exactly the same thing as you integrate over the P(v) function over v.

 

[T-7]

If you know the process is done in straight line in P-V diagram, to get the work done, just calculate the area under that straight line path. Can't get any easier. But anyway, that's exactly the same thing as you integrate over the P(v) function over v.

Sure. Thanks for confirming that. Regarding the heat, I presume using Cp or Cv is not sensible here, and I should just use an expression for the internal energy (U=5/2NkT, I think) and the 1st law to get it indirectly (U = dQ - dW, and we have just got dW, so we can state dQ). But perhaps this isn't the only way to do it?