When 0.668 L of Ar at 1.20 atm and 227°C is mixed with 0.258 L of O2 at 501 torr and 127°C in a 400 mL flask at 27°C, what is the pressure in the flask?

Pressure is given for both, am I supposed to use v1p1/t1 = v2p2/t2?

Some reason this really doesn't make sense to me. Help is seriously appreciated.

No, the (P1V1/T1) thing won't get it. You have two gases with P, T, and V given. I would calculate n for each using PV = nRT, add n for one to n for the other, then plug the total n back into a new PV = nRT where the new V is the 400 mL flask @ 27 C.

I can help you understand how to solve this problem! The ideal gas law equation, which you mentioned (v1p1/t1 = v2p2/t2), is indeed applicable here. Let's walk through the steps to solve this problem.

1. First, let's convert the temperatures from Celsius to Kelvin, as the ideal gas law requires temperature to be in Kelvin. To convert from Celsius to Kelvin, add 273.15 to the Celsius value.

Given temperatures:
Ar: 227°C = (227 + 273.15) K = 500.15 K
O2: 127°C = (127 + 273.15) K = 400.15 K
Flask: 27°C = (27 + 273.15) K = 300.15 K

2. Next, let's convert the pressures to a consistent unit. We can convert torr to atm by dividing by 760 (since 1 atm = 760 torr).

Given pressures:
Ar: 1.20 atm
O2: 501 torr ÷ 760 = 0.6579 atm

3. Now we can set up the equation using the ideal gas law:
(v1p1 / t1) = (v2p2 / t2)

We have the following values:
v1 (volume of Ar) = 0.668 L
p1 (pressure of Ar) = 1.20 atm
t1 (temperature of Ar) = 500.15 K
v2 (volume of O2) = 0.258 L
p2 (pressure of O2) = 0.6579 atm
t2 (temperature of flask) = 300.15 K

Plug in these values to the equation:
(0.668 L * 1.20 atm) / 500.15 K = (0.258 L * 0.6579 atm) / 300.15 K

4. Now solve the equation for the pressure in the flask:
p2 = (0.668 L * 1.20 atm * 300.15 K) / (500.15 K * 0.258 L)
p2 = 2.42 atm

So, the pressure in the flask is approximately 2.42 atm.