course Phy 202
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I am still working on this problem and I think I am on the right track but it is not working out just right. The problem is -- a monatomic gas in a 3.5 liter container is originally at 27 C and atmospheric pressure. It is heated at constant volume until its temperature is 180 C, then at constant pressure until the gas has increased its volume by .52 liters. How much thermal energy is required? By how much does the internal energy of the gas change? How much work is done in the process?
This is the work I did for the first part of the problem and you said it looked good. In the first part of the problem (volume is constant) Vo=3.5L, To=300K, Po=1atm and we know that temp changes to 453K. The first law of thermodynamics tells us that when V is contant W=0 and Q='dU. Since this is a monatomic gas I used 'dU=nCv'dt. One mol occupies 22.4L at stp so 3.5L is about .156mol.
I should have looked a little more closely at your calculation of n; the .156 was very reasonable so I didn't check your reasoning closely enough. You are right about 22.4 L at STP. But this system is not at the standard temperature, which is 273 K.
Use PV = n R T to find n. It will be a little less than the .156 mol you give here--about 10% less, or around .141 mol.
'dU=.156mol(12.465J/molK)(153K)=297.5J = Q.
For the second part you said to use 'dV and P to find Work--'dV=.52L(.00052m^3) and using PV=nRT I found P for the first part -- P=nRT/V=.156mol(8.21J/mol*K)(300K)/(.0035m^3)=109779.4Pa.
The expansion occurs after reaching 180 C, so the temperature is 453 K, not 300 K.
The number of moles needs to be corrected also, per my previous note.
This will give you a significantly higher pressure (circa 160 kPa), which will affect your result below.
Work=P'dV=.00052m^3*109779.4kg/m*s^2)=57.08J.
Next you said to use the ideal gas law to find the temp required to expand the gas to its new volume--
T=PV/nR=(109779.4kg/m*s^2)(.00052m^3)/(.179mol*8.21J/molK)=38.84K (the .179 mol came from the 3.5Liter+.52L=4.02L=.179mol).
V is the volume of the gas. Looks like you're using the change in volume, which doesn't belong in the gas law.
Use the new volume of the gas, the original plus the change. And you now know the pressure during the expansion (160 kPa give or take a lenient measure for my mental calculation).
The double-check this result. The temperature started at 453 K before expansion, and since P and n are constant the temperature ratio is equal to the volume ratio (T2 / T1 = V2 / V1). See if this gives you the same result you got from direct application of PV = n R T.
Next I found the energy required by using Cp=5/2R -- Q=mCp'dt=.179mol(20.525J/molK)(38.84K)=142.69J.
The number of moles isn't going to change, and I think the temperature change will be more than that (nearly but not quite double, but it's late and remember I'm calculating that mentally).
'dU=3/2R'dt=3/2(8.21J/molK)(38.84K)=535.39J/mol. Q should equal 'dU+'dW -- but I have 535.39J/mol for 'dU and 57.08J for W. I am not sure where I went wrong (Q is 297
You can't compare J/mol with J--they measure different things.
3/2 R is the constant-volume molar specific heat. To get Q, which would be in Joules, you would have to multiply this not only by the temperature change but the the number of moles.