# 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?

Jul 8, 2017

$P = 1.55$ $\text{atm}$

#### Explanation:

Here's how I go about doing this.

What we can do is use the ideal gas equation for both $\text{Ar}$ and ${\text{O}}_{2}$ 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 ($400$ $\text{mL}$) and temperature (${24}^{\text{o""C}}$).

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 $\text{mol Ar}$

n_ ("O"_2) = ((0.580cancel("atm"))(0.200cancel("L")))/((0.082057(cancel("L")•cancel("atm"))/("mol"•cancel("K")))(388cancel("K"))) = 0.00365 ${\text{mol O}}_{2}$

${n}_{\text{total}} = 0.0218$ $\text{mol Ar}$ $+ 0.00365$ ${\text{mol O}}_{2}$ $= 0.0254$ $\text{mol}$

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"