Internal Energy Thermodynamics

  • #1
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Homework Statement


CO2 is at P=3atm, T = 295K and V=1.2m3.
It is isobarically heated to T = 500K.
Find ΔU and ΔH

Homework Equations


dU = cpdT

The Attempt at a Solution


I am having a hard time in general in this class. I understand that in this problem, ΔP = 0. Does this mean that there must be a ΔV? Also, I am having a hard time understanding how to use cp if in this case, it changes with temperature. I do not think I need to use integration, but I am not sure.
 

Answers and Replies

  • #2
DrClaude
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I understand that in this problem, ΔP = 0. Does this mean that there must be a ΔV?
Yes.

Also, I am having a hard time understanding how to use cp if in this case, it changes with temperature. I do not think I need to use integration, but I am not sure.
If ##c_p## changes with temperature, then you need to integrate.
 
  • #3
Andrew Mason
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Homework Statement


CO2 is at P=3atm, T = 295K and V=1.2m3.
It is isobarically heated to T = 500K.
Find ΔU and ΔH

Homework Equations


dU = cpdT
No. dU = nCvdT

The Attempt at a Solution


I am having a hard time in general in this class. I understand that in this problem, ΔP = 0. Does this mean that there must be a ΔV?
CO2 can be treated as an ideal gas, so: PV = nRT. If T increases and P stays the same, what happens to V?
Also, I am having a hard time understanding how to use cp if in this case, it changes with temperature. I do not think I need to use integration, but I am not sure.
Cp is the heat capacity at constant pressure. So multiplying Cp by the change in temperature and number of moles gives you .......?. Since H = U + PV, can you determine the ΔH? (Hint: for constant pressure changes, how is it related to ΔQ?). Use ΔU = nCvΔT to determine change in internal energy.

AM
 
  • #4
BvU
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dU = cpdT wasn't that dH :smile: ?
Does this mean that there must be a ΔV?
Yes. Ideal gas law is good enough.
Also, I am having a hard time understanding how to use cp if in this case, it changes with temperature. I do not think I need to use integration
Either you integrate (stepwise), or you look it up. it indeed changes with T

I am having a hard time in general in this class
That's OK. With thermodynamics the sequence is: completely disoriented, then gradually more confident, and -- by the time you Master it -- complete disorientation again :smile:

[edit] wow, three responses !
 
  • #5
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Okay so I am familiar with the PV=nRT formula. In our class, we usually use PV=mRT, and use the specific R for the gas in question. Would my first step be to calculate the mass of CO2 in state 1 using this formula?
 
  • #6
Andrew Mason
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Okay so I am familiar with the PV=nRT formula. In our class, we usually use PV=mRT, and use the specific R for the gas in question. Would my first step be to calculate the mass of CO2 in state 1 using this formula?
Yes. Does the mass change?

AM
 
  • #7
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3
No I wouldn't think that mass would change. I went ahead and calculated the volume of the second state, although I'm not sure if it was needed. Would I then have to use dU = mcvdT ? And since it cv is not constant, would I need to integrate both sides of this equation?
 
  • #8
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I am confused because I do not understand which value to choose for Cp if it changes with volume. Even if I were to integrate I would have:
∫dU = m∫Cp dT
But how can I do that? The Cp is really throwing me off, and I have no idea how to use it in this problem. The same goes for Cv, since that also changes with temperature. Any help?
 
  • #9
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For an ideal gas, Cv and Cp are both functions only of temperature. They are related to the changes in internal energy and enthalpy by:
$$dU=nC_vdT$$
$$dH=nC_pdT$$
If the heat capacities are functions of temperature, then, to be exact, you need to integrate. But, often, if the temperature interval is small, the heat capacities can be approximated as being constant.
 
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