Question

A sample contains 4.80 grams of O2 gas has an initial volume of 15.0 L. If pressure and temperature remain constant: What is the final volume if 0.500 mole of O2 gas is added?

Answer 1

We calculate the molar volume of gas under given conditions:

$$n(O_2) = m/M = 4.80/32 = 0.15mol$$

$$V_m = V/n = 15.0/0.15 = 100 L/mol$$

Then, the amount of oxygen under given conditions:

$$V(O_2) = V_mn = 100*0.5 = 50.0L$$

Total:

$$V(O_2)_{total} = 50.0 +15.0 = 65.0L$$

Answer: 65.0L