# How many moles of oxygen are lost in the following problem?

## A cylinder with a moveable piston records a volume of 12.7 L when 3.5 mol of oxygen is added. The gas in the cylinder has a pressure of 5.83 atm. The cylinder develops a leak, and the volume of the gas is now recorded to be 12.1 L at the same pressure.

Oct 20, 2016

.16 moles of Oxygen assuming the temperature is at standard temperature.

#### Explanation:

Use Boyles Law to find the volume of Oxygen lost at standard pressure of 1 atm.

${P}_{1} \times {V}_{1} = {P}_{2} \times {V}_{2}$

${P}_{1} = 5.83$
${V}_{1} = 0.6$
${P}_{2} = 1.00$
${V}_{2}$ = Volume of Oxygen lost.

Solving the problem to find the Volume of Oxygen lost

$5.83 \times 0.6 = 1 \times {V}_{2}$

3.50 liters = ${V}_{2}$

22.4 liters = 1 mole at STP for an ideal gas.

Assuming Oxygen is close to an ideal gas

$\frac{3.50}{22.4}$ = moles Oxygen.

0.156 = moles of Oxygen.

Oct 22, 2016

The cylinder has lost 0.2 mol of oxygen.

#### Explanation:

The pressure and temperature are constant.

Only the volume and the number of moles are changing.

The appropriate relationship is therefore Avogadro's law: the volume of a gas is directly proportional to the number of moles, if all other variables are held constant.

In symbols,

V ∝ n

$V = k n$ or $\frac{V}{n} = k$

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {V}_{1} / {n}_{1} = {V}_{2} / {n}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$
${V}_{1} = \text{12.7 L"; n_1 = "3.5 mol}$
${V}_{2} = \text{12.1 L"; n_2 = "?}$
n_2 = n_1 × V_2/V_1 = "3.5 mol" × (12.1 color(red)(cancel(color(black)("L"))))/(12.7 color(red)(cancel(color(black)("L")))) = "3.33 mol"
Δn = n_2 - n_1 = "3.33 mol - 3.5 mol" = "-0.2 mol" (1 significant figure)