# 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?

##### 1 Answer

#### 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

#color(blue)(rho = m/V)#

#rho = "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

#color(blue)(P_1V_1 = P_2V_2)" "# , where

Plug in your values and solve for **do not** forget to convert the pressure of the gas from *mmHg* to *atm*

#P_1V_1 = P_2V_2 implies V_2 = P_1/P_2 * 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 = "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 = color(green)("11.4 g/L")#