At 85 degrees celsius and 760 torr, what are the amount of substance in the liquid and vapour phase when heptane is equals to 0.40?
This question cannot be answered without additional information. The amount of substance in the liquid and vapor phase depends on the total amount of heptane present, as well as its vapor pressure at the given temperature. The given information only specifies the temperature and total pressure, but not the amount of heptane present or its vapor pressure.
For a solution containing 1 mole each of hexane and heptane (a),estimate the vapour pressure in torr at 70 degrees celsius when vaporization on reduction just begins...(b).what is the vapour pressure of the solution at 70 degrees celsius when one drop of liquid remain, (c),estimate the mole fraction of hexane in liquid and vapour phase for the condition in (a).,(d), what are the mole fraction for the condition in (b)...(e),at 85 degrees celsius and 760 torr, what are the amount of substance in the liquid and vapour phase when hetpane is equals to 0.40?
(a) To estimate the vapor pressure of the solution at the point where vaporization on reduction just begins, we need to determine the mole fraction of the vapor. This can be estimated using Raoult's law, which states that the vapor pressure of a component in a solution is proportional to its mole fraction in the solution. At the point where vaporization on reduction just begins, the mole fraction of the vapor is equal to the mole fraction of the more volatile component in the liquid. In this case, hexane is more volatile than heptane, so the mole fraction of the vapor is equal to the mole fraction of hexane in the liquid.
Using Raoult's law, we can estimate the vapor pressure of each component in the solution as:
PHexane = XHexane * PHexane^o
PHeptane = XHeptane * PHeptane^o
where XHexane and XHeptane are the mole fractions of hexane and heptane in the liquid, and PHexane^o and PHeptane^o are the vapor pressures of pure hexane and heptane at 70 degrees Celsius, respectively.
At the point where vaporization on reduction just begins, the vapor pressure of the solution is equal to the sum of the partial pressures of hexane and heptane:
P = PHexane + PHeptane
Since the mole fraction of the vapor is equal to XHexane, we can rewrite this equation as:
P = XHexane * PHexane^o + (1 - XHexane) * PHeptane^o
We can solve for XHexane by setting P equal to the vapor pressure at the point where vaporization on reduction just begins. This pressure is not given in the question, so we cannot provide a numerical answer to this part.
(b) At the point where one drop of liquid remains, the vapor is saturated with respect to both hexane and heptane. This means that the mole fraction of each component in the vapor is equal to its mole fraction in the liquid. We can use Raoult's law as before to estimate the partial pressure of each component in the vapor, and then add them to get the total vapor pressure:
PHexane = XHexane * PHexane^o
PHeptane = XHeptane * PHeptane^o
P = PHexane + PHeptane
Since the vapor is saturated, we have:
XHexane = 1 - XHeptane
We can substitute this into the expressions for PHexane and PHeptane:
PHexane = (1 - XHeptane) * PHexane^o
PHeptane = XHeptane * PHeptane^o
and then substitute into the equation for P:
P = (1 - XHeptane) * PHexane^o + XHeptane * PHeptane^o
We can solve for XHeptane by setting P equal to the vapor pressure at the point where one drop of liquid remains. Again, this pressure is not given in the question, so we cannot provide a numerical answer to this part.
(c) At the point where vaporization on reduction just begins, we have:
XHexane (liquid) = XHexane (vapor)
This is because the vapor is in equilibrium with the liquid, and the mole fraction of the more volatile component in the vapor is equal to its mole fraction in the liquid. Since we don't know the vapor pressure at this point, we cannot provide a numerical answer, but XHexane (liquid) must be less than 0.5, since heptane is the majority component.
(d) At the point where one drop of liquid remains, we have:
XHexane (liquid) = XHeptane (liquid)
XHexane (vapor) = XHeptane (vapor)
This is because the vapor is saturated with respect to both components, so each component has the same mole fraction in the vapor as it does in the liquid. Since we don't know the vapor pressure at this point, we cannot provide a numerical answer, but XHexane (liquid) and XHeptane (liquid) must both be 0.5, since the solution is at its azeotropic composition.
(e) At 85 degrees Celsius and 760 torr, we cannot estimate the amount of substance in the liquid and vapor phases without additional information. We need to know the mole fraction of heptane in the liquid and the vapor to estimate the amounts of substance using the ideal gas law and the total pressure.
To find the amount of substance in the liquid and vapor phases of heptane at 85 degrees Celsius and 760 torr, we can use the concept of equilibrium vapor pressure.
The given value of 0.40 represents the equilibrium vapor pressure of heptane at the given temperature. This means that the vapor pressure of heptane in equilibrium with its liquid phase at 85 degrees Celsius is 0.40 atmospheres.
To calculate the amount of substance in each phase, we can use the Ideal Gas Law and the equation for vapor pressure. The Ideal Gas Law is expressed as:
PV = nRT
Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = amount of substance (in moles)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
Since we are given the pressure (760 torr = 1 atm) and do not have information about the volume, we will not consider the volume factor. Therefore, the equation becomes:
P = nRT
To find the amount of substance in the liquid phase, we need to calculate the liquid phase's pressure. The total pressure is given as 760 torr, which consists of the vapor pressure and the pressure due to the liquid phase. Therefore:
Total pressure (760 torr) = Vapor pressure (0.40 atm) + Pressure due to the liquid phase
Rearranging the equation, we can determine the pressure due to the liquid phase:
Pressure due to the liquid phase = Total pressure - Vapor pressure
Pressure due to the liquid phase = 760 torr - (0.40 atm * 760 torr) / 1 atm
Pressure due to the liquid phase = 760 torr - 304 torr
Pressure due to the liquid phase = 456 torr
Now, we can calculate the amount of substance in the liquid phase using the equation PV = nRT, substituting in the values we have:
Liquid phase pressure (456 torr) = nRT
n = (Liquid phase pressure) / (RT)
n = (456 torr) / [(0.0821 L·atm/(mol·K)) * (85 + 273.15) K]
n ≈ 0.0321 mol
Therefore, the amount of substance in the liquid phase is approximately 0.0321 moles.
Since we know the pressure and the amount of substance in the liquid phase, we can find the amount of substance in the vapor phase by subtracting the liquid phase amount from the total amount:
Amount of substance in the vapor phase = Total amount - Amount in liquid phase
Amount of substance in the vapor phase ≈ 0.40 mol - 0.0321 mol
Amount of substance in the vapor phase ≈ 0.3679 mol
Therefore, the amount of substance in the vapor phase is approximately 0.3679 moles.