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
Dec 18, 2015


#rho = "11.4 g/L"#


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

#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_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")#