What is the Internal Energy of 158 moles of CO2

AI Thread Summary
The discussion focuses on calculating the change in internal energy for 158 moles of carbon dioxide (CO2) cooled from 36°C to 25°C at constant pressure. The process is identified as isobaric, allowing for the determination of work done using the formula W = P(V1 - V0). The work calculated is approximately -14400 J. Participants discuss the use of the specific heat capacities (CP and CV) and the relationship between them through the ratio γ. The final calculation for the change in internal energy is confirmed to be nCvΔT.
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Homework Statement


A gas bottle contains exactly 158 moles of carbon dioxide CO2. Find the change in the internal energy of this much CO2 when it is cooled from 36C down to exactly 25C at a constant pressure of 1 atm. The gas can be treated as an ideal gas with γ=1.289. The gas constant reads R=8.314 J/(mol⋅K).

Homework Equations


PV=nRT,

W=∫PdV,

ΔU=Q-W,

The Attempt at a Solution


The gas is at constant pressure throughout, so this is an isobaric process. Meaning that volume and temperature change, which also means that there is work being done and there is heat flow.

I can easily get the work done;

PV0=nRT0 ⇒V0=(nRT0)/P

PV1=nRT1 ⇒V1=(nRT1)/P

So the work done,

W=∫PdV=P(V1-V0)

⇒ P((nRT1)/P-(nRT0)/P)

⇒ nR(T1-T1)

Substituting the required values gives,

W=(158 mol)×(8.314 J/(mol⋅K))*(25C-36C)=-14449.732 J

W=-14400 J to 3sf.

But I don't know where I am supposed to use γ to find the heat transferred, the only equations I can think of are;

dQ=nCPdT,

γ=CP/CV

CP=CV+R
 
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You have two equations in two unknowns (you already know γ and R). Just solve them for Cp and Cv.
 
Chestermiller said:
You have two equations in two unknowns (you already know γ and R). Just solve them for Cp and Cv.

Got it, thank you very much.
 
As a check on your answer, the change in internal energy should be nCvΔT.
 
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