Answer:
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
