How much butane is present as a liquid?
given: Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 degrees celsius. A 250 mL sealed flask contains 0.55g of butane at - 22 degrees celsius.
Answer:
You get the vapor pressure of butane by the Clausius-Clapeyron relation for an ideal gas phase:
dlnpv / dT = ΔHv / (R·T²)
If integrate the relation from the normal boiling point
given with T₀= -0.4°C = 272.75K and pv₀= 1atm = 101325Pa
and solve for pv, you get:
ln (pv(T)/pv₀) = (-ΔHv/R)· (1/T - 1/T₀)
=>
pv(T) = pv₀· exp{ -(ΔHv/R)· (1/T - 1/T₀) }
at T = -22°C = 251.15K
pv = 101325Pa · exp{ -(22440J/mol / 8.314472J/molK)· (1/251.15K - 1/272.75) }
= 43263Pa
Assume liquid volume in the flask is negligible. Apply ideal gas law
p · V = m · Rb · T
(Rb = R / Mb = 8.314472J/molK / 58.08g/mol = 0.143155J/gK is the specific gas constant of butane)
If there is liquid in the flask the pressure of the gaseous butane equals the vapor pressure. Thus the mass of gaseous butane is:
m_g = (pv · V) / (Rb · T)
= (43263Pa · 0.25·10^-3m³) / ( 0.143155J/gK · 251.15K)
= 0.3g
The liquid mass in the flask is
m_l = 0.55g - m_g = 0.25g
The liquid volume is
V_l = m_l / ρ_l = 0.25g / 0.584 g/cm3 = 0.43cm³
which is much smaller than the flask volume of 250cm³. So the assumption of negligible liquid volume was correct.
More Questions and Answers:
What are to clear liquids that when mixed give a bright color of anykind?
If 15.0 g ethyl alcohol (CH3CH2OH) are dissolved in 500 g water, find the molality of the solution?
Organic Chem Questions?
Why is it that no matter what color bubble bath you use the bubbles are always white?
All free elements have an oxidation number of...?
For which of the following acids would a 1.00 M solution have the lowest pH ( highest H+ concentration)?
Density problem?
I feel dumb.. how do I find moles? and a few other things.. !!?
Can you answer this?
Answer:
You get the vapor pressure of butane by the Clausius-Clapeyron relation for an ideal gas phase:
dlnpv / dT = ΔHv / (R·T²)
If integrate the relation from the normal boiling point
given with T₀= -0.4°C = 272.75K and pv₀= 1atm = 101325Pa
and solve for pv, you get:
ln (pv(T)/pv₀) = (-ΔHv/R)· (1/T - 1/T₀)
=>
pv(T) = pv₀· exp{ -(ΔHv/R)· (1/T - 1/T₀) }
at T = -22°C = 251.15K
pv = 101325Pa · exp{ -(22440J/mol / 8.314472J/molK)· (1/251.15K - 1/272.75) }
= 43263Pa
Assume liquid volume in the flask is negligible. Apply ideal gas law
p · V = m · Rb · T
(Rb = R / Mb = 8.314472J/molK / 58.08g/mol = 0.143155J/gK is the specific gas constant of butane)
If there is liquid in the flask the pressure of the gaseous butane equals the vapor pressure. Thus the mass of gaseous butane is:
m_g = (pv · V) / (Rb · T)
= (43263Pa · 0.25·10^-3m³) / ( 0.143155J/gK · 251.15K)
= 0.3g
The liquid mass in the flask is
m_l = 0.55g - m_g = 0.25g
The liquid volume is
V_l = m_l / ρ_l = 0.25g / 0.584 g/cm3 = 0.43cm³
which is much smaller than the flask volume of 250cm³. So the assumption of negligible liquid volume was correct.
The answers post by the user, for information only, FunQA.com does not guarantee the right.
More Questions and Answers: