7.00 L of argon gas is known to contain 78 grams. If the amount of gas is increased to 96 grams, what is the new volume (at an unchanged temperature and pressure)?

Question

please explain how you got the answer, thank you

Answer 1

If the temperature and pressure remain constant (as in the current problem), then adding gas must increase the volume. If not, then the pressure would certainly increase. If you look at the ideal gas law, PV = nRT, you see that if the T and P are constant, ( and R is also a constant), then V is proportional to n (moles of gas). Thus, we can set up a proportion ...

V1/n1 = V2/n2

V1 = initial volume = 7.00 L

n1 = initial number of moles = 78 g Ar x 1 mol/40 g = 1.95 moles

V2 = final volume = ?

n2 = final number of moles = 96 g Ar x 1 mol/40 g = 2.4 moles

7.00 L/1.95 moles = V2/2.4 moles

V2 = 8.62 L

NOTE: the number of moles increased by 2.4/1.95 = 1.23 x. The volume also increased by 8.62 L/7.00 L = 1.23 x. A direct relationship.