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December 20, 2014

December 20, 2014

Posted by **Nancy** on Tuesday, July 23, 2013 at 1:52am.

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.

Answer this Question

- Chemistry re-post -
**bobpursley**, Tuesday, July 23, 2013 at 10:43amHCL 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 re-post -
**DrBob222**, Tuesday, July 23, 2013 at 2:30pmFollowing 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 re-post -
**Devron**, Thursday, July 25, 2013 at 1:37amOkay 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.

**Answer this Question**

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