# 1.00 liter of O_2 gas at 760 torr pressure and 0°C has a mass of 1.42 g. If the pressure on the gas is increased to 8.00 atm at 0°C, what is the density of the O_2 in grams/liter?

Dec 18, 2015

$\rho = \text{11.4 g/L}$

#### Explanation:

Before doing any calculation, try to predict what you should see happen to the density of the gas.

Under the initial conditions for pressure and temperature, the oxygen gas had a density of

$\textcolor{b l u e}{\rho = \frac{m}{V}}$

$\rho = \text{1.42 g"/"1.00 L" = "1.42 g/L}$

Notice that temperature remains constant, but pressure increases. As you know, pressure and volume have an inverse relationship when temperature and number of moles of gas are kept constant - this is known as Boyle's Law.

So, if pressure increases, that means that volume will decrease. Since the same mass of gas will now occupy a smaller volume, you can say that the density of the gas will increase.

Now let's prove this by doing some calculations. The equation that describes Boyle's Law looks like this

$\textcolor{b l u e}{{P}_{1} {V}_{1} = {P}_{2} {V}_{2}} \text{ }$, where

${P}_{1}$, ${V}_{1}$ - the pressure and volume of the gas at an initial state
${P}_{2}$, ${V}_{2}$ - the pressure and volume of the gas at a final state

Plug in your values and solve for ${V}_{2}$ - do not forget to convert the pressure of the gas from mmHg to atm

${P}_{1} {V}_{1} = {P}_{2} {V}_{2} \implies {V}_{2} = {P}_{1} / {P}_{2} \cdot {V}_{1}$

V_2 = (760/760 color(red)(cancel(color(black)("atm"))))/(8.00color(red)(cancel(color(black)("atm")))) * "1.00 L" = "0.125 L"

This time, the density of the sample will be

$\rho = \text{1.42 g"/"0.125 L" = "11.36 g/L}$

I'll leave the answer rounded to three sig figs, despite the fact that you only have two sig figs for the initial pressure of the gas

$\rho = \textcolor{g r e e n}{\text{11.4 g/L}}$