A rigid, 2.50 L bottle contains 0.458 mol He. The pressure of the gas inside the bottle is 1.83 atm. If 0.713 mol Ar is added to the bottle and the pressure increases to 2.05 atm, what is the change in temperature of the gas mixture?The initial temperature of the gas is ______ K.

Answers

Answer 1

Answer: These are inert gases, so we can assume they don't react with one another. Because the two gases are also subject to all the same conditions, we can pretend there's only "one" gas, of which we have 0.458+0.713=1.171 moles total. Now we can use PV=nRT to solve for what we want.

The initial temperature and the change in temperature. You can find the initial temperature easily using PV=nRT and the information provided in the question (before Ar is added) and solving for T.

You can use PV=nRT again after Ar is added to solve for T, which will give you the final temperature. The difference between the initial and final temperatures is the change. When you're solving just be careful with the units!

SIDE NOTE: If you want to solve for change in temperature right away, you can do it in one step. Rearrange both PV=nRT equations to solve for T, then subtract the first (initial, i) from the second (final, f):

PiVi=niRTi --> Ti=(PiVi)/(niR)

PfVf=nfRTf --> Tf=(PfVf)/(nfR)

ΔT=Tf-Ti=(PfVf)/(nfR)-(PiVi)/(niR)=(V/R)(Pf/nf-Pi/ni)

In that last step I just made it easier by factoring out the V/R since V and R are the same for the initial and final conditions.

Answer 2

Answer:

A rigid, 2.50 L bottle contains 0.458 mol He. The pressure of the gas inside the bottle is 1.83 atm. If 0.713 mol Ar is added to the bottle and the pressure increases to 2.05 atm, what is the change in temperature of the gas mixture?

The initial temperature of the gas is

wrong 137

⇒ 122 K. correct on E.D.G

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