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swat32
3 years ago
12

Chamber 1 and Chamber 2 have equal volumes of 1.0L and are assumed to be rigid containers. The chambers are connected by a valve

that is initially closed. Chamber 1 contains 2.00 moles of helium and Chamber 2 contains 1.00 mol of helium. Both chambers are at a temperature of 27°C.Part 3.When the valve is opened, what happens to the pressure in Chamber 1? Choose the best answer.
Chemistry
1 answer:
vitfil [10]3 years ago
5 0
1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas  than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)

=> p1 = 2 p2

Which is easy to demonstrate using ideal gas equation:

p1 = nRT/V = 2.0 mol * RT / 1 liter

p2 = nRT/V = 1.0 mol * RT / 1 liter

=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2

2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.

So, the pressure in both chambers (which form one same vessel) is:

p = nRT/V = 3.0 mol * RT / 2liter

which compared to the initial pressure in chamber 1, p1, is:

p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1

So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.

You can also see how the pressure in chamber 2 changes:

p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
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Decigram is larger unit while milligram is smaller unit. One decigram is equal to the 100 milligram. In order to convert the given value into  decigram we have to divide the value by 100.

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4 0
4 years ago
For a concentration technique, Stotal is given as 19.31 ± 0.035, Smb is 0.22 ± 0.008, and kA is 0.154 ± 0.007 ppm–1, where Stota
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3 years ago
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