Wednesday Quiz #2 (2015)

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Sabrina Smelser 3D
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Joined: Fri Sep 26, 2014 2:02 pm

Wednesday Quiz #2 (2015)

Postby Sabrina Smelser 3D » Sun Feb 08, 2015 12:42 am

The question was as follows:

A 100.0L vessel containing 6.00mol of Ne at 2.00atm is cooled to 203.0K. Assuming the monoatomic gas behaves ideally, calculate delta U and delta H for this process and describe the thermodynamic reaction.

I calculated the initial temperature and got -203.2K but I am uncertain of which formulas to use from here.

Thank you!

Kayla Denton 1A
Posts: 106
Joined: Fri Sep 26, 2014 2:02 pm

Re: Wednesday Quiz #2 (2015)

Postby Kayla Denton 1A » Sun Feb 08, 2015 9:20 am

At constant pressure, delta H = q (heat).
You can find q from q = nC(delta T).
We know that n = 6.00 moles.
And C, since it is a monatomic ideal gas at constant pressure, must be 5/2R, where R is the gas constant.
What is the change in temperature? We must find the initial temperature and then do final - initial.
Temperature can be found from PV = nRT. We isolate for T so T = PV/nR. Make sure to check units for this part!!!
T = (2.00 atm)(100.0 L)/((6.00 mol)(0.08206 Latm/Kmol)). Make sure you use the right gas constant!
The answer is T initial = 406.2 K
We know that T final is 203.0 K
Thus the change in temperature is 203.0 - 206.2 = -203.2 K (like you correctly calculated).

Now plug all of that in to find q, thus delta H! Once again, make sure to use the right gas constant.
q = (6.00 mol)(5/2 x 8.3145 J/Kmol)(-203.2) = -25343 J rounded to -25.3 kJ. This is delta H.

Now for delta U. Fortunately, as well as being at constant pressure, we are also at constant volume. Therefore, delta U = q. Delta U also = -25.3 kJ.

At constant volume AND constant pressure, delta H = delta U!

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

Re: Wednesday Quiz #2 (2015)

Postby Niharika Reddy 1D » Sun Feb 08, 2015 9:56 am

My TA said that for ΔU, we would use C=3/2 R since we are at constant volume and plug that into q=nCΔT. Since constant volume means w=-PΔV=0, it follows that ΔU=q+w=q+0=q.

In this sense, ΔU and ΔH will not be the same, since Cv is used for calculating ΔU and Cp for ΔH.

Kayla Denton 1A
Posts: 106
Joined: Fri Sep 26, 2014 2:02 pm

Re: Wednesday Quiz #2 (2015)

Postby Kayla Denton 1A » Sun Feb 08, 2015 12:11 pm

Oh you're right, my bad! Delta U = delta H at constant volume and pressure only when we're working with solids and liquids. For gases, the gas constant changes. Sorry about that :)


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