Question

After 0.600 L of Ar at 1.47 atm and 221 degrees Celsius is mixed with 0.200 L of O2 at 441 torr and 115 degrees Celsius in a 400.-mL flask at 24 degrees Celsius, what is the pressure in the flask?

Answer 1

`P = 1.55` `"atm"`

Explanation:

Here's how I go about doing this.

What we can do is use the ideal gas equation for both `"Ar"` and `"O"_2` to find the total moles. Then we use the ideal gas equation a third time to find the total pressure with known total moles, volume (`400` `"mL"`) and temperature (`24^"o""C"`).

I won't show the unit conversions here, as I predict you already know how; the moles of each substance is

`n_ "Ar" = ((1.47cancel("atm"))(0.600cancel("L")))/((0.082057(cancel("L")•cancel("atm"))/("mol"•cancel("K")))(494cancel("K"))) = 0.0218` `"mol Ar"`

`n_ ("O"_2) = ((0.580cancel("atm"))(0.200cancel("L")))/((0.082057(cancel("L")•cancel("atm"))/("mol"•cancel("K")))(388cancel("K"))) = 0.00365` `"mol O"_2`

`n_"total" = 0.0218` `"mol Ar"` `+ 0.00365` `"mol O"_2` `= 0.0254` `"mol"`

We'll now use the ideal gas equation to solve for the pressure at the new conditions:

`P = (nRT)/V = ((0.0254cancel("mol"))(0.082057(cancel("L")•"atm")/(cancel("mol")•cancel("K")))(297cancel("K")))/(0.400cancel("L"))`

`= color(blue)(1.55` `color(blue)("atm"`