a ridged 2.40L sealed vessel contains He and C3H6 gases. The partial pressure of He is 1.4 atm and that of C3H6 is 1.7 atm, at 55 degress celcius. If the vessel is cooled to - 45 degress celcius, what is the partial pressure of each gas

Partial pressure He → 0.96 atm

Partial pressure C₃H₆ → 1.18 atm

Explanation:

We apply the Charles Gay Lussac law, to solve this. Pressure varies directly proportional to absolute T°, when the volume keeps on constant.

P₁ / T₁ = P₂ / T₂

We convert the T° to absolute T°

55°C + 273 = 328K

-45°C + 273 = 228K

Total pressure = Sum of partial pressures

1.7 atm + 1.4 atm = 3.1 atm

When we apply the formula we would know the new total pressure

3.1 atm / 328K = P₂ / 228K

(3.1 atm / 328K). 228K = 2.15 atm

As the moles has not been modified with the change of T°, we assume the mole fraction is still the same.

Mole fraction He = Partial pressure He / Total pressure

1.4 atm / 3.1 atm = 0.45

Mole fraction C₃H₆ = Partial pressure C₃H₆ / Total pressure

1.7 atm / 3.1 atm = 0.55

0.45 = Partial pressure He / 2.15 atm

Partial pressure = 0.45. 2.15 atm → 0.96 atm

0.55 = Partial pressure C₃H₆ / 2.15 atm

Partial pressure = 0.55. 2.15 atm → 1.18 atm