A balloon contains 0.121 mol of gas and has a volume of 2.73 L.

If an additional 7.2×10^−2 mol of gas is added to the balloon, what will its final volume be?

Vfinal=_________ L

.121/2.73 = (.121 + 7.2 x 10^-2)/x

.0443 = .193/x
.0443x = .193
x = 4.36 L

you use the formula PV=nRT, so rearrange for V so you get nRT/P

first solve for final volume by adding the moles together:

.072moles + .121 moles=.193moles (you add the two moles because it says they add that much gas)
.08206L atm/K mol is the gas constant R
Temperature is at 0 degrees Celsius so converted to kelvins its (0+273)=273K
Pressure is at 1 atm unless otherwise noted

V=(.193mol)(.08206L atm/K mol)(273K)
V=4.324L, 4.324L is the final volume