If 0.750 L of argon at 1.50 atm and 177oC and 0.235 L of sulfur dioxide at 95.0 kPa and 63.0oC are added to a 1.00 L flask and the flask's temperature is adjusted to 25.0oC

Question

If 0.750 L of argon at 1.50 atm and 177oC and 0.235 L of sulfur dioxide at 95.0 kPa and 63.0oC are added to a 1.00 L flask and the flask's temperature is adjusted to 25.0oC, what is the resulting pressure in the flask?

0.244 atm

0.0851 atm

1.74 atm

0.946 atm

Answer 1

Find number of moles of each from ideal gas law: n(Ar) = PV/RT = (1.50 atm)(0.750 L) / [(0.08206 L atm/mol K)(177 +273K)] = 0.03047 mole. n(SO₂) = (95.0 kPa)(0.235 L) / [(8.314 L kPa/mol K)(63 +273K)] = 0.00799 mole. Total moles = 0.03846. Total P = nRT/V = (0.03846 mol)(0.08206 L atm/mol K)(25 +273K) / (1.00 L) = 0.940 atm. So the last value 0.946 atm is closest.