Thermodynamics Gases: Final Temp, Work, Internal Energy, Q

In summary, the conversation involves compressing two cubic feet of air from 85°F and 13.9 psi to 115 psi. The final temperature, work, internal energy change (ΔU), and heat (Q) are requested. The equations for the first law of thermodynamics and ideal gas law are mentioned, and it is clarified that the process is adiabatic. The questioner is unsure how to approach the problem and asks for help, and the responder suggests using the ideal gas law to find the number of moles and the specific equations for adiabatic work. The questioner also asks about handling units and converting between grams and gram-moles, and the responder provides guidance on converting
  • #1
Mrencko
109
0

Homework Statement


two cubic ft, of air at 85 f and 13.9 psi, compress to 115 psi.
What is the final temp, the work, the internal energy "delta u", and Q

Homework Equations


well i don't have any equations, i am asking for the equations

The Attempt at a Solution


i don't know how to approach to this problem because i don't know if this is
Isothermal
or adiabatic process,
i am sufing the web, but there is a ton of equations out there and i feel overwhelmed about it, so i coul use some help to make a solid base to start in thermodynamics
 
Last edited:
Physics news on Phys.org
  • #2
Isobaric means "at constant pressure" - can you compress something in an isobaric process?
 
  • #3
sorry i made a mistake it was
Isothermal
 
  • #4
If it was isothermal - does it make sense to ask about the final temperature?
 
  • #5
Mrencko said:

Homework Equations


well i don't have any equations, i am asking for the equations
I think they meant for you to assume that the compression is adiabatic and reversible, although this certainly isn't clear from the problem statement. Have you learned the equation for the first law of thermodynamics? If so, why didn't you list that as one of your equations? What about the ideal gas law; did you not learn about that?
 
  • #6
Yes i know these equations, the ideal gas is pv=nrt and firts law of thermpfynamics its w= Q-internal energy
 
  • #7
you are right, it was so obvious, sorry its adiabatic because everything changes
how i should use this kind of info "one lb of air"? i don't see any weight or mass unit in equations, just "n" but if its air how i can know the moles of "air" i can if its two gases or one, for this problem i made the calculatins to find temperature final, and volume final, but i am stuck at the work calculation i have all exept the "n"

i convert all to non english units

V1=58.6337 lt
T1=302 k
P1=0.9458389 ATM
P2=7.8252 ATM
equations***
P1V1γ=P2V2γ
T1V1γ-1=T2V2γ-1

using that i found****
T2=552.34039 K
V2=12.9617 LT
now i just need the W and internal energy because the problem ask for Q but in adiabatic Q=0 so i need a litle litle help

i have this formulas for work***

W=-ncvλT→W=-ncv(T2-T1)
W=-(Cv)/(R)(P2V2-P1V1)
whic one its correct? and how i can know n?
 
Last edited:
  • #8
To get n, use PV=nRT together with the initial conditions.
 
  • #9
Mrencko said:
i have this formulas for work***

W=-ncvλT→W=-ncv(T2-T1)
W=-(Cv)/(R)(P2V2-P1V1)
whic one its correct?
Both equations are correct, and both should give the same answer.
 
  • #10
should i use R=(0.0820574)ltatm/kmol?
and what p and v, should work for p1v1=nRT1 and for p2v2=nRT2, i mean those two equations are valids or just whit initial valules, also if W=-internal energy any value of work its equal to a loss of internal energy and there is no need to calculate it_? also another question what if the problem says "1 lb of air" how i should consider the mass in equations?
 
Last edited:
  • #11
Mrencko said:
should i use R=(0.0820574)ltatm/kmol?
I'll leave it to you to work this out, considering the units that are needed.
and what p and v, should work for p1v1=nRT1 and for p2v2=nRT2, i mean those two equations are valids or just whit initial valules,
Try both equations, and see what you get.
also if W=-internal energy any value of work its equal to a loss of internal energy and there is no need to calculate it_?
The first equation you gave actually is that change in internal energy.
also another question what if the problem says "1 lb of air" how i should consider the mass in equations?
I don't understand this question. Are you asking how to convert lbs to kg, grams, or gram moles?
 
  • #12
Well my other problem are the "same" exept for the mass thing, this is an example:
"one diesel machine have 159.09 grams of air, compress from 14.7 psi to 465 psi, whith 23.8 c of initial temp and 525.55 of final temp, calculate w and Q" its the same but the grams or lb, thing i don't know how to handle in the calculations
 
  • #13
Mrencko said:
Well my other problem are the "same" exept for the mass thing, this is an example:
"one diesel machine have 159.09 grams of air, compress from 14.7 psi to 465 psi, whith 23.8 c of initial temp and 525.55 of final temp, calculate w and Q" its the same but the grams or lb, thing i don't know how to handle in the calculations
You don't know how to convert from grans to gram-moles, correct? The molecular weight of a material is equal to the number of grams per gram mole of material. So, to convert from grams to gram-moles, you divide by the molecular weight.
 
  • #14
thanks for the response, but i still have doubts, the "air its a mix" i get the idea of gramss/mols of x given gas or two, tree but what gases i should consider for air and proportions?
 
  • #15
Mrencko said:
thanks for the response, but i still have doubts, the "air its a mix" i get the idea of gramss/mols of x given gas or two, tree but what gases i should consider for air and proportions?
Use 21 mole percent for oxygen and 79 mole percent for nitrogen. This gives a molecular weight of (0.21)(32)+(0.79)(28)=29
 
  • #16
thank you very much i was really piss of about this because the teacher dint explain nothing at all, many thanks
 

Related to Thermodynamics Gases: Final Temp, Work, Internal Energy, Q

1. What is the final temperature of a gas after an isobaric process?

The final temperature of a gas after an isobaric process can be calculated using the formula Tf = Ti + (Q / nCp), where Tf is the final temperature, Ti is the initial temperature, Q is the heat added to the system, n is the number of moles of gas, and Cp is the specific heat at constant pressure.

2. How is work calculated in a thermodynamics gas process?

The work done in a thermodynamics gas process can be calculated using the formula W = PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume of the gas.

3. What is internal energy in thermodynamics gases?

Internal energy in thermodynamics gases refers to the total energy of the gas particles, including both kinetic and potential energy. It is affected by changes in temperature, pressure, and volume of the gas.

4. How is heat (Q) calculated in a thermodynamics gas process?

Heat in a thermodynamics gas process can be calculated using the formula Q = nCpΔT, where Q is the heat added to the system, n is the number of moles of gas, Cp is the specific heat at constant pressure, and ΔT is the change in temperature.

5. What is the relationship between internal energy, work, and heat in a thermodynamics gas process?

In a thermodynamics gas process, the change in internal energy (ΔU) is equal to the sum of the work (W) and heat (Q) added to the system, or ΔU = Q + W. This is known as the First Law of Thermodynamics.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
691
  • Introductory Physics Homework Help
Replies
1
Views
933
  • Biology and Chemistry Homework Help
Replies
2
Views
2K
  • Classical Physics
Replies
3
Views
1K
  • Biology and Chemistry Homework Help
Replies
3
Views
5K
Replies
13
Views
2K
  • Materials and Chemical Engineering
Replies
1
Views
993
  • Introductory Physics Homework Help
Replies
3
Views
990
  • Introductory Physics Homework Help
Replies
10
Views
1K
Replies
12
Views
2K
Back
Top