The stopcock connecting a 4.87 L bulb containing hydrogen gas at a pressure of 4.38 atm, and a 4.87 L bulb containing argon gas

Question

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The stopcock connecting a 4.87 L bulb containing hydrogen gas at a pressure of 4.38 atm, and a 4.87 L bulb containing argon gas

at a pressure of 2.90 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is _____atm

Chemistry

1 answer:

andrew-mc [135] · Answer 1 · 2022-07-30T14:09:08+03:00

Answer:

P = 3,63 atm

Explanation:

Using PV=nRT and knowing the temperature in the problem remains constant it is possible to rewrite this formula, thus:

PV = kn, where k is RT (Constant), P is pressure, V is volume and n are moles.

The first bulb contains:

4,87L×4,38atm = 21,3k moles

And the second bulb:

4,87L×2,90atm = 14,1k moles

When gases are mixed, the total moles are:

21,3k moles + 14,1k moles = 35,4k moles

As total volume is 4,87L + 4,87L = 9,74L

Replacing:

P = kn/V

P = 35,4k moles / 9,74L

P = 3,63 atm

I hope it helps!