H2Otown_WX Posted September 23, 2012 Share Posted September 23, 2012 I'm having some trouble with a thermodynamics problem which asks what the change in internal energy (dU) of dry air is in an isothermal process where T = 7 deg.C and the change in specific volume is 100 cm^3/g. Here's what I figured out: From the 1st Law: dH = dU + dW where dH is an increment of heat added and dW is the work done, Pd-alpha (alpha representing specific volume). Since specific heat c = dH/dT, I concluded that dH must be 0 given that dH = c*dT = 0. What is tripping me up is that no pressure was given to calculate dW. From what I read online, if air is treated as an ideal gas, the change in internal energy would be 0 because no heat is flowing into or out of the air during expansion because it is at constant temperature. This wasn't stated anywhere in my book so I just wanted a second opinion. Link to comment Share on other sites More sharing options...
ohleary Posted September 23, 2012 Share Posted September 23, 2012 Have you seen this: http://en.wikipedia.org/wiki/Isothermal_process Link to comment Share on other sites More sharing options...
H2Otown_WX Posted September 23, 2012 Author Share Posted September 23, 2012 Have you seen this: http://en.wikipedia....thermal_process Yeah lol. That's what I was referring to here: "From what I read online, if air is treated as an ideal gas, the change in internal energy would be 0 because no heat is flowing into or out of the air during expansion because it is at constant temperature." Link to comment Share on other sites More sharing options...
ohleary Posted September 23, 2012 Share Posted September 23, 2012 Yeah lol. That's what I was referring to here: "From what I read online, if air is treated as an ideal gas, the change in internal energy would be 0 because no heat is flowing into or out of the air during expansion because it is at constant temperature." Well, specifically I meant this: Where you don't need pressure to solve for dW. If dH = dU + dW and dH = 0, then dU = -dU dU = -nRTln Vb/Va Maybe not, it's been years and years since I've looked at this stuff! Link to comment Share on other sites More sharing options...
H2Otown_WX Posted September 24, 2012 Author Share Posted September 24, 2012 Well, specifically I meant this: Where you don't need pressure to solve for dW. If dH = dU + dW and dH = 0, then dU = -dU dU = -nRTln Vb/Va Maybe not, it's been years and years since I've looked at this stuff! Interesting...I think you may very well be right. Link to comment Share on other sites More sharing options...
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