Thermodynamic Processes

You can also write it this way: dU=C_V\,dT+T(\partial P/\partial T)_V\,dV. So, for an ideal gas, (\partial P/\partial T)_V=R/V, and you can solve for C_V pretty easily. Sound good?
  • #1
Texag
20
0

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.
 
Physics news on Phys.org
  • #2
Texag said:
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: [itex]dU=C_V\,dT[/itex] for a constant-volume process, and [itex]dH=C_P\,dT[/itex] for a constant-pressure process. But for an ideal gas, [itex]U=C_VT+U_0[/itex]; the energy depends on temperature only. So for an ideal gas, [itex]\Delta U=C_V\Delta T=(C_P-R)\Delta T[/itex] for all processes; the [itex]C_V[/itex] (or [itex]C_P[/itex]) is just a parameter. Sound good?
 
  • #3
Mapes said:
Hold on: [itex]dU=C_V\,dT[/itex] for a constant-volume process, and [itex]dH=C_P\,dT[/itex] for a constant-pressure process. But for an ideal gas, [itex]U=C_VT+U_0[/itex]; the energy depends on temperature only. So for an ideal gas, [itex]\Delta U=C_V\Delta T=(C_P-R)\Delta T[/itex] for all processes; the [itex]C_V[/itex] (or [itex]C_P[/itex]) is just a parameter. Sound good?

Thanks for the reply. So, is it just when not under the ideal gas assumption that [itex]dU=C_V\,dT[/itex] 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.
 
  • #4
For any closed, constant-volume system, [itex]dU=C_V\,dT[/itex]. This comes from the definition of heat capacity [itex]C_X=T(\partial S/\partial T)_X[/itex], where X is the constraint condition (like constant volume) and the general differential energy expression [itex]dU=T\,dS-P\,dV+\mu\,dN[/itex] where the change in volume [itex]dV=0[/itex] and the change in matter [itex]dN=0[/itex].
 

What is a thermodynamic process?

A thermodynamic process is a physical change that occurs in a system due to the transfer or conversion of energy. It can involve changes in temperature, pressure, volume, and other properties of the system.

What are the three types of thermodynamic processes?

The three types of thermodynamic processes are isobaric (constant pressure), isochoric (constant volume), and adiabatic (no heat exchange). These processes can occur in different combinations and can be described using thermodynamic equations and diagrams.

What is the first law of thermodynamics?

The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, only transferred or converted from one form to another. This law is the basis for understanding and analyzing thermodynamic processes.

What is the difference between reversible and irreversible processes?

A reversible process is one in which the system can be returned to its original state by reversing the steps of the process. In contrast, an irreversible process cannot be reversed and results in a permanent change in the system. Reversible processes are ideal and often used for theoretical calculations, while irreversible processes occur in real-world systems.

What is thermodynamic equilibrium?

Thermodynamic equilibrium occurs when a system and its surroundings are in a state of balance, with no net energy transfer or change in properties. In other words, the system is stable and no further changes will occur unless an external force is applied. Systems strive to reach thermodynamic equilibrium, and many thermodynamic processes are driven by this tendency.

Similar threads

  • Advanced Physics Homework Help
Replies
5
Views
934
  • Classical Physics
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
14
Views
7K
Replies
1
Views
497
  • Introductory Physics Homework Help
Replies
19
Views
1K
Replies
3
Views
2K
  • Advanced Physics Homework Help
Replies
4
Views
9K
Replies
22
Views
2K
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
113
Back
Top