Butane has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 degrees C. A 250mL sealed flask contains 0.5g of butane at 25 degrees C.
How much butane is present as a liquid?
If the butane is warmed to 25 degrees C, how much is present as a liquid
For the second part, use the Clausius-Clapeyron equation.
P butane at normal boiling point will be 760 torr. Solve for P @ 298, then substitute P into PV = nRT and solve for n. Compare with the number of moles (0.5 g/molar mass butane) present in liquid form.
For part a, I think you can use PV = nRT as is (P at -0.4 C is 1 atm), then compare n from this with n of 0.5 g butane.
To determine how much butane is present as a liquid in the sealed flask, we need to compare the current temperature (25 degrees C) with the normal boiling point (-0.4 degrees C).
At temperatures below the normal boiling point, a substance will exist primarily as a liquid. At temperatures equal to or above the boiling point, the substance will exist primarily as a gas.
Let's calculate the amount of butane present as a liquid at 25 degrees C:
1. Convert the mass of butane from grams to moles:
Given mass = 0.5 g
Molar mass of butane (C4H10) = 58.12 g/mol
Moles of butane = mass / molar mass
Moles of butane = 0.5 g / 58.12 g/mol
2. Calculate the number of moles that remain as a liquid:
Since the boiling point of butane is below the current temperature, all of the butane will exist as a liquid at 25 degrees C.
Therefore, the entire 0.5 g of butane will be present as a liquid in the sealed flask at 25 degrees C.
To determine how much butane is present as a liquid in the given scenario, we need to compare the boiling point of butane with the temperature at which it is present. Since the normal boiling point of butane is -0.4 degrees C, any temperature above this value will cause it to vaporize.
Now, let's calculate the number of moles of butane in the 250 mL sealed flask containing 0.5g of butane at 25 degrees C.
First, convert the mass of butane to moles using the molar mass of butane. The molar mass of butane (C₄H₁₀) is 58.12 g/mol.
moles of butane = mass of butane / molar mass of butane
moles of butane = 0.5g / 58.12 g/mol
moles of butane ≈ 0.0086 mol
Next, we need to calculate the volume of the liquid butane when it is at its boiling point. This can be done using the ideal gas law:
PV = nRT
Where:
P = pressure (assumed to be constant at the boiling point)
V = volume (liquid butane)
n = number of moles
R = ideal gas constant
T = temperature (boiling point of butane)
Since the system is sealed, the pressure remains constant, and we can use the equation in simplified form:
V = nRT / P
Given that the boiling point of butane is -0.4 degrees C, we need to convert this temperature to Kelvin:
T = -0.4 + 273.15 = 272.75 K
Assuming a constant pressure of 1 atmosphere, the ideal gas constant (R) is 0.0821 L·atm/(mol·K).
V = (0.0086 mol)(0.0821 L·atm/(mol·K))(272.75 K) / (1 atm)
V ≈ 0.193 L
Therefore, when the butane is at its boiling point, approximately 0.193 L (or 193 mL) of butane will be present as a liquid.
Now, let's determine how much butane will be present as a liquid when it is warmed to 25 degrees C.
Since the temperature is above the boiling point, all of the butane will be present as a gas at this temperature. Thus, at 25 degrees C, none of the butane will be present as a liquid.