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?
3 answers
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.
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.