# How do you solve for V2, using Avogadro's Law when moles of a gas are added(n2)? A sample containing 7.20g O2 gas has a volume of 35.0L. Pressure and temperature remain constant. What is the new volume if 0.800mole O2 gas is added?

Jun 24, 2014

The new volume is 159 L.

#### Explanation:

n_1 = 7.20 color(red)(cancel(color(black)("g O"_2))) × ("1 mol O"_2)/(32.00 color(red)(cancel(color(black)("g O"_2)))) = "0.225 mol O"_2

So we have

${V}_{1} = \text{35.0 L}$; ${n}_{1} = \text{0.225 mol}$

If we add ${\text{0.800 mol O}}_{2}$

${V}_{2} = \text{?}$; ${n}_{2} = \text{0.225 mol + 0.800 mol" = "1.025 mol}$

V_2 = V_1 × n_2/n_1

${V}_{2} = \text{35.0 L" × (1.025 color(red)(cancel(color(black)("mol"))))/(0.225 color(red)(cancel(color(black)("mol")))) = "159 L}$

This makes sense.

The number of moles increased by about fivefold, so the volume should have increased about five-fold (to about 175 L).