Question

Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 ∘C. A 250 mL sealed flask contains 0.8 g of butane at −22∘C.
How much butane is present as a liquid?

Answer 1

Given that:

• The heat of vaporization = 22.44 kJ/mol = 22440 J/mol
• normal boiling point which is the initial temperature = 0.4° C = (273 + (-0.4))K  = 272.6 K
• volume  = 250 mL = 0.250 L
• Mass of butane = 0.8 g
• the final temperature = -22° C = (273 + (-22)) K = 251 K

The first step is to determine the vapor pressure at the final temperature of 251K by using the Clausius-Clapeyron equation. This is following by using the ideal gas equation to determine the numbers of moles of butane gas. After that, the mass of butane present in the liquid is determined by using the relation for the number of moles.

Using Clausius-Clapeyron Equation:

$\mathbf{In (\dfrac{P_2}{P_1} )= -\dfrac{\Delta H_{vap}}{R}(\dfrac{1}{T_2} - \dfrac{1}{T_1})}$

where;

P1 and P2 correspond to the temperature at T1 and T2.

∴

replacing the values into the given equation, we have;

$\mathbf{In \dfrac{P_2}{1\ atm} = -\dfrac{22440 \ J/mol}{8.314 \ J/mol.K}(\dfrac{1}{251 \ K} - \dfrac{1}{272.6 \ K})}$

$\mathbf{In \dfrac{P_2}{1\ atm} =-(0.852053785)}$

$\mathbf{P_2=0.427 \ atm}$

As such, at -22° C; the vapor pressure = 0.427 atm

Now, using the ideal gas equation:

PV = nRT

where:

• P = Pressure
• V = volume
• n = number of moles of butane
• R = universal gas constant
• T = temperature

∴

Making (n) the subject of the formula:

$\mathbf{n = \dfrac{PV}{RT}}$

$\mathbf{n = \dfrac{0.427 atm \times 0.250 L}{(0.08206 \ L.atm/k.mol) \times 251}}$

$\mathbf{n =0.00518 mol}$

We all know that the standard molecular weight of butane = 58.12 g/mol

∴

Using the relation for the number of moles which is:

$\mathbf{number \ of \ moles = \dfrac{mass}{molar mass}}$

mass = 0.00518 mole × 58.12 g/mol

mass = 0.301 g

∴

The mass of butane in the flask = 0.301 g

But the mass of the butane present as a liquid in the flask is

= 0.8 g - 0.301 g

= 0.499 g

In conclusion, the mass of the butane present as a liquid in the flask is 0.499 g

Learn more about vapourization here:

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