Why Is Cv Used in Adiabatic Processes?

[Texag]

Homework Statement

A gas in a piston-cylinder assembly undergoes a thermodynamic cycle consisting of three processes.
Process 1-2: adiabatic compression with pV1.4 = constant from p*1 = 50 psia and V1 = 3 ft3 to V2 = 1 ft3.
Process 2-3: isometric,
Process 3-1: isobaric with U*1 - U*3 = 46.7 Btu.
Neglect kinetic and potential energy changes. Show the pV diagram of the cycle with arrows indicating work & heat transfer during each process labeled according to the statements above. Show the direction of each process with arrows. Show all work on back of the page as needed. Determine
a. The net work.
b. The heat transfer for process 3-1 in Btu.
c. The temperature T1, if this is air as an ideal gas.
d. The change in temperature, T1 – T3, if cv = constant.

Am I correct in assuming that the mass should be provided?

Homework Equations

dU= dQ-dW

pV=RT

dW=p*dV

dH = dU + d(p*V)

The Attempt at a Solution

Found: parts a and b without trouble.

Having trouble finding T1 with given information.

I also am confused with the use of Cp and Cv in general. I know dU=Cp*dT for isobaric processes, and dU=Cv*dT for isometric processes. I do not understand why Cv should be used for an adiabatic process, when it is obviously not isometric or isobaric. I also do not understand the fact that Enthalpy and Internal Energy are seemingly interchangeable for Cp, i.e. dH=dU=Cp*dT for isobar.

I'd really appreciate some help with this, thanks guys.

 

[Mapes]

I also am confused with the use of Cp and Cv in general. I know dU=Cp*dT for isobaric processes, and dU=Cv*dT for isometric processes. I do not understand why Cv should be used for an adiabatic process, when it is obviously not isometric or isobaric.

Hold on: $dU=C_V\,dT$ for a constant-volume process, and $dH=C_P\,dT$ for a constant-pressure process. But for an ideal gas, $U=C_VT+U_0$; the energy depends on temperature only. So for an ideal gas, $\Delta U=C_V\Delta T=(C_P-R)\Delta T$ for all processes; the $C_V$ (or $C_P$) is just a parameter. Sound good?

 

[Texag]

Hold on: $dU=C_V\,dT$ for a constant-volume process, and $dH=C_P\,dT$ for a constant-pressure process. But for an ideal gas, $U=C_VT+U_0$; the energy depends on temperature only. So for an ideal gas, $\Delta U=C_V\Delta T=(C_P-R)\Delta T$ for all processes; the $C_V$ (or $C_P$) is just a parameter. Sound good?

Thanks for the reply. So, is it just when not under the ideal gas assumption that $dU=C_V\,dT$ for a constant-volume process (only)? Just seemed a little redundant, but if I was missing out on assuming ideal gas, then that explains it.

Thanks again.

 

[Mapes]

For any closed, constant-volume system, $dU=C_V\,dT$. This comes from the definition of heat capacity $C_X=T(\partial S/\partial T)_X$, where X is the constraint condition (like constant volume) and the general differential energy expression $dU=T\,dS-P\,dV+\mu\,dN$ where the change in volume $dV=0$ and the change in matter $dN=0$.