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 0.5 g of butane at -22 C.
How much butane is present as a liquid?
If the butane is warmed to 25 C, how much is present as a liquid?
-22C is well below the boiling point. Only a small amount will be present as vapor. You can estimate that amount using the Clausius-Clapeyron equation and the 1 atm vapor pressure at -0.4 C.
You can also use the C-C equation for the vapor pressure at the higher temperature, 25 C. It might all be vapor but start by finding out how much of the mass can be in the vapor state in that volume at that temperature.
You can also look up the vapor pressures at both temperatures, but they probably want you to calculate them from the information provided.
A graph of log of vapor pressure vs 1/T is usually a very straight line.
Use the Clausius-Clapeyron equation to calculate pressure at -22. Then PV = nRT should get n for the gas and you can go from there.
0.204 gm
Well, let's do some math to figure this out!
To determine how much butane is present as a liquid at -22°C, we need to compare the initial temperature (-22°C) to the boiling point (-0.4°C). Since -22°C is below the boiling point, we can assume that all the butane will be in a liquid state. So, 0.5 grams of butane is present as a liquid.
Now, if we warm up the butane to 25°C, which is above the boiling point, some of it will start to vaporize. To determine how much remains as a liquid, we need to calculate the amount of heat required to vaporize the butane.
First, let's convert 0.5 grams of butane to moles. The molar mass of butane (C4H10) is approximately 58.12 g/mol, so:
0.5 g / 58.12 g/mol = 0.0086 mol
Now, we can calculate the heat required to vaporize this amount of butane:
0.0086 mol * 22.44 kJ/mol = 0.193 kJ
So, if we warm up the butane to 25°C, approximately 0.193 kJ worth of butane will vaporize, and the rest will remain as a liquid.
Keep in mind that these calculations assume ideal conditions, and there might be some small fluctuations due to factors like pressure and impurities. But hey, that's the joy of chemistry!
To determine how much butane is present as a liquid, we need to compare the temperature of the butane with its boiling point.
First, let's calculate the number of moles of butane present in the flask at -22 °C.
Step 1: Convert mass to moles
The molar mass of butane (C4H10) can be calculated as follows:
4(C) + 10(H) = 4(12.01 g/mol) + 10(1.01 g/mol) = 58.12 g/mol
The number of moles can be calculated using the formula:
moles = mass (g) / molar mass (g/mol)
For 0.5 g of butane:
moles = 0.5 g / 58.12 g/mol ≈ 0.0086 mol
Now let's determine the amount of butane that exists as a liquid at -22 °C.
Step 2: Calculate the boiling point in Kelvin
To convert the boiling point from Celsius to Kelvin, we add 273.15 to the value:
-0.4 °C + 273.15 = 272.75 K
Step 3: Use the Clausius-Clapeyron equation
The Clausius-Clapeyron equation relates the vapor pressure of a substance to its temperature and heat of vaporization. The equation is as follows:
ln(P₂/P₁) = (-ΔHvap/R) * (1/T₂ - 1/T₁)
Where:
P₁ = vapor pressure at temperature T₁
P₂ = vapor pressure at temperature T₂
ΔHvap = heat of vaporization
R = gas constant (8.314 J/(mol·K))
In this case, we can assume the vapor pressure at -22 °C is close to zero, so the equation simplifies to:
ln(P₂) = -ΔHvap/R * (1/T₂ - 1/T₁)
For T₁ = 272.75 K and T₂ = -22 °C = 251.15 K, the equation becomes:
ln(P₂) = -ΔHvap/R * (1/251.15 K - 1/272.75 K)
Step 4: Solve for P₂
First, let's convert the heat of vaporization from kJ/mol to J/mol:
ΔHvap = 22.44 kJ/mol * 1000 J/kJ ≈ 22,440 J/mol
Substituting the values into the equation:
ln(P₂) = -(22,440 J/mol) / (8.314 J/(mol·K)) * (1/251.15 K - 1/272.75 K)
Calculating this equation will give us the natural logarithm of the vapor pressure at -22 °C. To solve for P₂, we take the exponential of both sides:
P₂ = e^(ln(P₂))
Now we can calculate P₂.
Step 5: Calculate the vapor pressure at -22 °C
Using a scientific calculator or math software, compute the expression:
P₂ = e^(-22,440 J/mol) / (8.314 J/(mol·K)) * (1/251.15 K - 1/272.75 K)
The resulting value of P₂ will be the vapor pressure at -22 °C.
Step 6: Determine the liquid butane present
To determine how much butane is present as a liquid at -22 °C, we compare the given pressure to the vapor pressure at that temperature. If the given pressure is lower, then all the butane is in the liquid state.
If the pressure at -22 °C is lower than the vapor pressure, the entire 0.5 g of butane will be present as a liquid.
To calculate the amount of butane present as a liquid if warmed to 25 °C, we need to repeat steps 2 to 6 for the new temperature.
Note: Additional information is needed to calculate the vapor pressure at the given temperatures, such as the Antoine equation constants for butane or tabulated vapor pressure data. These calculations assume ideal behavior, but real gases can deviate from ideal behavior at high pressures or low temperatures.
a) is around 2.0
b) is zero, all the substance has vaporized at 25 degress there for none of it will be left as a liquid