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Constant Pressure Process

  • Thread starter VitaX
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  • #1
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Homework Statement



Suppose 15.2 g of oxygen (O2) is heated at constant atmospheric pressure from 13.3°C to 148°C. (a) How many moles of oxygen are present? (Take the molar mass of oxygen to be 32.0 g/mol) (b) How much energy is transferred to the oxygen as heat? (The molecules rotate but do not oscillate.) (c) What fraction of the heat is used to raise the internal energy of the oxygen?

Homework Equations



Diatomic gas Cp = (7/2)R ; Cv = (5/2)R
Q = nCp(Tf - Ti)
W = P(Vf - Vi) = NR(Tf - Ti)
Change in Kinetic Energy = (3/2)nRT
Change in Internal Energy = nCv(Tf - Ti) = Q - W
Cp - Cv = R

The Attempt at a Solution



I believe I got all the information correct above. The ones I'm not too sure about are the Cp and Cv values.

Part A
15.2 g O2 (1 mol O2/32 g O2) = .475 mol O2

Part B
Change in Internal Energy = (.475 mol O2)((5/2)*8.31 J/mol*K)(148 C - 13.3 C) = 1329.236 J

Part C
Change in Kinetic Energy = (3/2)(.475 mol O2)(8.31 J/mol*K)(148 C - 13.3 C) =797.542 J

Fraction of heat used to raise Internal Energy of O2 = 797.542 J/1329.236 J = 0.6

To be honest I'm not really sure about my work on Parts B and C. I was kind of guessing as I was going along. Is the work for those parts correct?
 

Answers and Replies

  • #2
ehild
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For B, calculate the heat. For C, find the ratio between change of internal energy and heat. You do not need the translational kinetic energy.

ehild
 
  • #3
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For B, calculate the heat. For C, find the ratio between change of internal energy and heat. You do not need the translational kinetic energy.

ehild
Part B
Q = nCpΔT = n(7/2*R)ΔT = (.475 mol O2)(29.085 J/mol*K)(148 C - 13.3 C) = 1860.931 J

Part C
ΔEint = nCvΔT = n(5/2*R)ΔT = (.475 mol O2)(20.775 J/mol*K)(148 C - 13.3 C) = 1329.236 J

Ratio = 1329.236 J/1860.931 J = 0.7143

How's this look? I'm not sure if the ratio is correct or not. Should be it Q/ΔEint instead?
 
Last edited:
  • #4
ehild
Homework Helper
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Some heat is taken and the gas does work, so only a fraction of the heat is converted to internal energy (rises the temperature). You do not need even to evaluate the heat and internal energy to get this fraction, as the heat is Q=7/2 n R ΔT and ΔEint=5/2 n R ΔT, so (ΔEint)/Q =5/7 .

ehild
 
  • #5
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Some heat is taken and the gas does work, so only a fraction of the heat is converted to internal energy (rises the temperature). You do not need even to evaluate the heat and internal energy to get this fraction, as the heat is Q=7/2 n R ΔT and ΔEint=5/2 n R ΔT, so (ΔEint)/Q =5/7 .

ehild
Ah, yeah you are right. I guess finding ΔEint would only help reassure you of that fact to see if your calculations are correct.

On a side note though, how do you know when to use Q equation and when to use ΔEint? That's kind of tricking me up. In the question it asked "How much energy is transferred to the oxygen as heat?" Since it said heat at the end is that what gives it away to use the equation for Q? And what does it mean exactly when it says the molecules rotate but do not oscillate? Is that really useful information here?
 
  • #6
ehild
Homework Helper
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The first law of thermodynamics states that the internal energy can be increased either by adding heat to the gas or doing work on the gas ΔEint=Q+W. The energy transferred to the gas as heat is simply Q. The energy transferred to the gas as work is W. If it is a reversible process, that is the external pressure and the internal one are in equilibrium (or very near to equilibrium) the work done on the gas is equal to the negative work of the gas (-pΔV for a constant pressure process). In this case the First Law is written as ΔEint=Q-W(gas) or Q=ΔEint+W(gas). For an equilibrium process, all equations are equivalent, you only need to take care if you use the work done by the gas or the work done on the gas by external forces.

You know that the internal energy of an ideal gas is f/2 n R T, where n is the degree of freedom. It is 3/2 for a monoatomic gas, where the particles have only translational degrees of freedom. For a diatomic molecule, there are two additional rotational degrees of freedom, and for a three-atomic one, there are 3. At higher temperatures, the vibrations of the molecule are excited. If there are N atoms in the molecule it has 3N degrees of freedom altogether. 3 belongs to the translation of the centre os mass. 3 or 2 degrees belong to the rotation. The others are vibrational degrees of freedom, but they contribute to the internal energy only at high temperatures.

ehild
 
  • #7
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DPT = Displacement of pressure theory

The idea that movement through the field of medium becomes the pressure dynamics that create form, i.e. vortexes.
 

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