I posted this before and am sorry to post again and take up room but it is due this week and I'd like to understand how to get the answer (no one has responded yet):
question
The nuclide 38cl decays by beta emission with a half life of 40min. A sample of .40 m 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?
2 answers
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.
I got the correct answer on this problem. The problem you are having is with the partial pressures. You have .400 mol H38Cl at the start. The 38Cl undergoes 2 half lives to .100 mol. So now in the solution you have left .100 mol of H38Cl. The H gas that falls off of this would become H2 gas in that state. So you have .300 mol H fall off of the H38Cl. That combines to be .150 mol H2. The beta decay of the 38Cl resulted in .300 mol 38Ar. The remaining amounts gives you .55 mol of gas. Apply that to PV=nRT. Don't screw up the units.
2 answers