Question

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? Use the correct number of significant digits.
The initial temperature of the gas is

Answer 1

Please show work.

Answer 2

Hello!

Since the pressure is fairly low (<5 atm) and you are especially dealing with inert gases that have minimal tendency to react, you can apply the ideal gas law.

\(PV=nRT\)
You must first find the initial temperature before argon is added
\((1.83~atm)(2.50~L)=(0.458~mol~He)(R)(T)\)
Since the pressure is in atmospheres, the R-constant you will use would be \(0.08206~\frac{L•atm}{mol•K}\)
\((1.83~atm)(2.50~L)=(0.458~mol~He)(0.08206~\frac{L•atm}{mol•K})(T_i)\)
\(T_i=121.729~K\)
I do not suggest you round to s.f. now as it may result in errors later.

Now 0.713 moles of argon are being added to the bottle. Since these are inert gases, in conjunction to the KMT, you can conclude that the mols of gas are additive, and that there is no reaction. Hence, the new mols of gas are
\(0.458~mol~He+0.713~mol~Ar=1.171~mol~gas\)
Now you must use PV=nRT again
\((2.05~atm)(2.50~L)=(1.171~mols)(0.08206~\frac{L•atm}{mol•K})(T_f)\)
\(T_f=53.334~K\)

\(∆T=T_f-T_i=53.33~K-121.729~K\)

You could also use the combined gas law to solve this problem;
\(\frac{P_1V_1}{n_1T_1}=\frac{P_2V_2}{n_2T_2}\)

Cheers!