A sample containing 4.80 g of #O_2# gas has a volume of 15.0 L at constant pressure and temperature. What is the new volume, in liters, after 0.500 mole of #O_2# gas is added to the initial sample of 4.80 g of #O_2#?

1 Answer
Feb 16, 2017

Answer:

The new volume is 65.0 L.

Explanation:

Initially, you have

#4.80color(red)(cancel(color(black)("g O"_2))) × "1 mol O"_2/(32.00 color(red)(cancel(color(black)("g O"_2)))) = "0.1500 mol O"_2#

Thus, you have

#"0.1500 mol O"_2/"15.0 L" = "0.0100 mol O"_2/"1 L"#

If you now add 0.500 mol of #"O"_2#, you will have

#"(0.1500 + 0.500) mol O"_2 = "0.6500 mol O"_2#

Since volume is directly proportional to the number of moles, the new volume is

#0.6500 color(red)(cancel(color(black)("mol O"_2))) × "1 L"/(0.0100 color(red)(cancel(color(black)("mol O"_2)))) = "65.0 L"#