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

• Mrencko
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

## 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
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

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Isobaric means "at constant pressure" - can you compress something in an isobaric process?

sorry i made a mistake it was
Isothermal

If it was isothermal - does it make sense to ask about the final temperature?

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?

Yes i know these equations, the ideal gas is pv=nrt and firts law of thermpfynamics its w= Q-internal energy

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?

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To get n, use PV=nRT together with the initial conditions.

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.

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?

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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?

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

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.

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?

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

thank you very much i was really piss of about this because the teacher dint explain nothing at all, many thanks