Posted by Nancy on Tuesday, July 23, 2013 at 1:52am.
I took a look at this question to see if I copied it correctly, and it does NOT say how much of the solution was used. It does use mols as the unit.
Dr. Bob, I am not sure what you mean by "decays into Argon." I realize that the 38 CL decays to Argon, but how would that work into the solution? I am confused as it appears I need to use PV =NRT and then also half life, but I just don't get 300K as the answer, which is what the book has.
If you have any insight, let me know. If not I will find out later this week how to solve it.
Thanks!
question
The nuclide 38cl decays by beta emission with a half life of 40min. A sample of .40 mol of H38cl is placed in a 6.24 liter container. After 80 min the pressure is 1650 mmHg. What is temperature of container?
I have tried calculating the decay rate of the 38cl and then plugging into the formula pv =nrt but I do not get the answer, 300 k. Any help would be great.
I would think it would be like this:
.40 mol is .0258% Hydrogen
.40 mole is .974 CL38
.40 x .974=.3896 x half life .25=.0974 + mole of hydrogen .01032
=.1077
13.54/.1077= 125k but this isn't correct
The correct answer is 300K
anybody know how it was obtained?
Chemistry  DrBob222, Monday, July 22, 2013 at 12:39pm
I'm responding just to let you know that I've looked at the problem more than once over the last day or two. Is the HCl a gas? I suppose no if it is a 0.40 m solution. Is that m and not M? How much of the 0.40m solution? is used? The problem doesn't say. And do you realize that Cl38 goes to Ar38 upon releasing a beta particle.
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Chemistry repost  bobpursley, Tuesday, July 23, 2013 at 10:43am
HCL is a gas. There is a trick her.
2Cl2 decays to Cl2+2Ar
so note that for each two moles of CL2 decaying, you get three moles of gas as products. That may solve your problem

Chemistry repost  DrBob222, Tuesday, July 23, 2013 at 2:30pm
Following Bob Pursley's comment, IF HCl is a gas (then what does the m dstand for?)
k = 0.693/t_{1/2} = about 0.017 but you can do it more accurately.
ln(No/N) = kt
ln(0.4/N) = 0.0173*80
N = about 0.1
So we started at 0.4
end up with 0.1
used = 0.3 mol
0.3 x 3/2 = 0.45 gas formed and we had 0.1 left over which makes 0.55 mols gas.
Then PV = nRT and T = 300 K.

Chemistry repost  Devron, Thursday, July 25, 2013 at 1:37am
Okay guys, this problem is really bugging me. I initially saw it and declined to try and solve it. However, after Nancy reposted the question, I thought that I would see if I could take a stab at it. I did back calculations and obtained 0.55 moles, but I could not figure out how in the world that a closed system, which the wording of the problem indicates that it is, undergoing beta decay could produce more moles then was initially present before the decay process. I calculated that 0.1 moles of H38Cl gas would remain in the system after 80 minutes, and I figured that H38Cl is a gas, but I have never seen a problem where a substance decays and produces more moles of gas then was initially present. From what I can tell, the law of conservation states that the end process of a decay or reaction can never exceed that starting material. However, in this case it did. I'm just trying to understand the initial 1:1.5 ratio; I understand what Dr. Bob222 did, but not the initial step. A clearer explanation would be appreciated.
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