Answer : The vapor pressure of propane at is 17.73 atm.
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of propane at
= ?
= vapor pressure of propane at normal boiling point = 1 atm
= temperature of propane =
= normal boiling point of propane =
= heat of vaporization = 24.54 kJ/mole = 24540 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the vapor pressure of propane at is 17.73 atm.
Answer:
1255.4L
Explanation:
Given parameters:
P₁ = 928kpa
T₁ = 129°C
V₁ = 569L
P₂ = 319kpa
T₂ = 32°C
Unknown:
V₂ = ?
Solution:
The combined gas law application to this problem can help us solve it. It is mathematically expressed as;
P, V and T are pressure, volume and temperature
where 1 and 2 are initial and final states.
Now,
take the units to the appropriate ones;
kpa to atm, °C to K
P₂ = 319kpa in atm gives 3.15atm
P₁ = 928kpa gives 9.16atm
T₂ = 32°C gives 273 + 32 = 305K
T₁ = 129°C gives 129 + 273 = 402K
Input the values in the equation and solve for V₂;
V₂ = 1255.4L