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 $O_2$?

Md Ariful Islam

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Md Ariful Islam

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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

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

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