If 0.750 L of argon at 1.50 atm and 177°C and 0.235 L of sulfur dioxide at 95.0 kPa and 63.0°C are added to a

Question

1.00-L flask and the flask's temperature is adjusted to 25.0°C, what is the resulting pressure in the flask?

Answer 1

From the ideal gas law:
PV = nrT

Keep in mind that nr will be constant...so: P1 * V1 / T1 = P2 * V2 / T2.

You can set this up as:
Pa * Va / Ta + Ps * Vs / Ts = Pf * Vf / Tf

Why? The argon and the sulfur dioxide essentially become one gas mixture sharing pressure, volume, and temperature in the flask at the end.

First, you need to calculate the pressure of the sulfur dioxide into atmospheres: 95 kPa = .937577 atm.
Second, the temperatures need to be in Kelvin, so add 273.15 to all temperatures.

1.5 atm * .75 L / 450.15 K + .937577 atm * .235 L / 336.15 = Pf * 1 L / 298.15 K
(0.002499 + .0006555) * 298.15 = Pf
Pf = 0.0031545 * 298.15 = 0.941 atm (solution)