Thermodynamics - Ideal Diatomic Gas


by ChronicQuantumAddict
Tags: diatomic, ideal, thermodynamics
ChronicQuantumAddict
ChronicQuantumAddict is offline
#1
Nov6-05, 02:21 AM
P: 41
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?
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
ChronicQuantumAddict
ChronicQuantumAddict is offline
#2
Nov6-05, 10:37 PM
P: 41
anyone please?
Andrew Mason
Andrew Mason is offline
#3
Nov6-05, 11:20 PM
Sci Advisor
HW Helper
P: 6,572
Quote Quote by ChronicQuantumAddict
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?
The first one is fairly straightforward. Think PV=nRT. If T is constant, what does that tell youi about PV at all times? That should give you Vf. What is U at intial and final volumes? What is W in terms of P and [itex]\Delta V[/itex]? That should give you W. What is the relationship between U, Q and W? That will give you Q.

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

ChronicQuantumAddict
ChronicQuantumAddict is offline
#4
Nov8-05, 06:19 PM
P: 41
Unhappy

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?
ChronicQuantumAddict
ChronicQuantumAddict is offline
#5
Nov8-05, 09:33 PM
P: 41
please help, need to turn this in tomorrow and im stuck
Andrew Mason
Andrew Mason is offline
#6
Nov8-05, 10:32 PM
Sci Advisor
HW Helper
P: 6,572
Quote Quote by ChronicQuantumAddict
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?
U is the internal energy due to kinetic energy of the molecules. If you keep the volume constant and increase the temperature, the gas does no work. So the increase in internal energy is equal to the heat input. Since the heat flow is just the heat capacity of the gas at constant volume multiplied by the change in temperature, [itex]dU = nC_vdT => U = \int dU = \int nC_vdt = nC_vT[/itex].

So [itex]U = nC_vT[/itex] (T in K).

AM
ChronicQuantumAddict
ChronicQuantumAddict is offline
#7
Nov9-05, 04:22 PM
P: 41
understood now, thanks very much AM


Register to reply

Related Discussions
a diatomic ideal gas quesion Introductory Physics Homework 0
Thermodynamics/Ideal gas Introductory Physics Homework 2
Questions about the ideal gas law - thermodynamics Introductory Physics Homework 0
Ideal Gas Law + 1st Law of Thermodynamics Introductory Physics Homework 9
Ideal Gas: Equation Of State in Diatomic Gases: still valid? Classical Physics 11