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Offline elfishies

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Enthalpy problem
« on: February 09, 2016, 05:09:25 AM »
Consider a sample containing 2.00 mol of a monatomic ideal gas that undergoes the following changes from State A to State D:
State A PA=12.5atm, VA=14.00L
State B PB=12.5atm, VB=5.00L
State C PC=22.5atm, VC=5.00L
State D PD=22.5atm, VD=21.50L
Assume that the external pressure is constant during each step and equals the final pressure of the gas for that step.
Q: What is the value of q for the overall change from State A (through States B and C) to State D?


I'm lost and don't know where to start, can anyone tell me what formula to use? I know that q=deltaE+PdeltaV but i'm not sure if that's the right formula to use here, if it is how do I calculate deltaE? thanks!

Offline mjc123

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Re: Enthalpy problem
« Reply #1 on: February 09, 2016, 07:08:26 AM »
As all the steps are either at constant pressure or constant volume, I think the easiest way is
1. Calculate T for each state using the ideal gas law
2. For isovolumetric steps, q = 2*Cv,m*ΔT
3. For isobaric steps, q = 2*Cp,m*ΔT
Of course you know what Cv,m and Cp,m are for a monatomic ideal gas?

Offline elfishies

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Re: Enthalpy problem
« Reply #2 on: February 09, 2016, 05:59:54 PM »
I'm not sure if i'm doing this right

First i used PV=nRT --> T=PV/nR
state A: T=(12.5atm)(14L)/(2mol)(8.3145JK-1mol-1)=10.524K
state B: T=3.758
state C: T=6.765
state D: T=29.091

state A to B: q=n(5R/2)(delta T)--> q=2((5)(8.3145)/2)(3.758-10.524)=-281.249
B to C: q=2((3)(8.3145)/2)(6.765-3.758)=74.999
C to D: q=928.125

add them together and it equals 722 kJ?


Offline mjc123

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Re: Enthalpy problem
« Reply #3 on: February 10, 2016, 04:04:31 AM »
If you're using P in atm and V in L, what units does R need to be in?

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