Question

A sample containing 4.80 g of O2 gas has an initial volume of 15.0 L. What is the final volume, in liters, when each of the following occurs and pressure and temperature do not change?
A) a sample of 0.500 mole of O2 is added to the 4.80g of O2 in the container

B) a sample of 2.00 g of O2 is removed

C) A sample of 4.00 g of O2 is added to the 4.80 g of O2 gas in the container

Answer 1

A) Ideal gas law is

PV = nRT

V = nRT/P

If temperature and pressure are constant , Volume in depends on no of mole , n

V1 = n1 k

V2 = n2 k

V1/V2 = n1/n2

V2 = n2 × V1/n1

initial mole n1 = 4.80g/32g/mol =0.15mol

initial volume , V1 = 15L

final volume = 0.15mol + 0.500mol = 0.65mol

Therefore,

Final volume,V2 = 15L × 0.65mol/0.15mol = 65L

B) no of mole of O2 removed = 2g/32g/mol = 0.0625mol

final mole of O2, n2 = 0.15mole - 0.0625mole = 0.0875mol

Final volume V2 = 15L × 0.0875mol/ 0.15mol = 8.75L

C) no of mole of O2 added = 4g/32g/mol = 0.125

final mol = 0.15 + 0.125 = 0.275mol

Final volume , V2 = 15L × 0.275mol/0.15mol = 27.5L