Question

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`?

Answer 1

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"`