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Thermodynamics - Ideal Diatomic Gas |
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| Nov6-05, 02:21 AM | #1 |
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Thermodynamics - Ideal Diatomic Gas
The question is as follows:
An ideal diatomic gas for which cv = (3/5)*R, where R is the ideal gas constant, and occupies a volume V= 2 meters^3, at a Pressure Pi= 5 atm and temperature Ti = 20 degrees Celcius (= 293 K). The gas is compressed to a final pressure of Pf = 12 atm. Calculate the final volume Vf, the final temperature Tf, the work done dW, the heat transferred dQ, and the change in Internal energy dU for the following: a). a reversible isothermal compression b). a reversible adiabatic compression i know in the adiabatic process, dQ = 0, and in the isothermal dT = 0. Where do i take it from here?
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| Nov6-05, 10:37 PM | #2 |
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anyone please?
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| Nov6-05, 11:20 PM | #3 |
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Recognitions:
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For the second part, it is more complicated. You have to use the adiabatic relationship: [tex]PV^\gamma = constant[/tex] where [itex]\gamma = C_p/C_v[/itex] What is [itex]\gamma[/itex] here? What is the expression for W? How is W related to Q and U? AM |
| Nov8-05, 06:19 PM | #4 |
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Thermodynamics - Ideal Diatomic Gas i am stuck with the internal energy for part a) isothermal process.here is what i've done so far. since its isothermal: P1*V1 = P2*V2 and thus,V2= (P1/P2)*V1 = 0.8333333 m^3 For isothermal, Ti = Tf = T = constant = 293.15 K W = -nRT*(Integral: Pi to Pf) (dP/P) = -nRT*ln(P2/P1) = nRT*ln(P1/P2) thus W = -2.134 * 10^6 J, work done on the system Now, for U i am stuck. i keep reading in the book, that for an ideal gas U only depends on T, and they give me this formula. U - U_0 = dU = Cv*dT, where Cv is the specific heat at constant volume. but volume is not constant in this problem, and also, this leads to an integral with dT and for isothermal dT = 0 If this is true, then the integral yields a constant U and that means U_final = U_initial, but that still doesnt tell me what U is. Anything further? where am i going wrong? |
| Nov8-05, 09:33 PM | #5 |
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please help, need to turn this in tomorrow and im stuck
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| Nov8-05, 10:32 PM | #6 |
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Recognitions:
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So [itex]U = nC_vT[/itex] (T in K). AM |
| Nov9-05, 04:22 PM | #7 |
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understood now, thanks very much AM
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