The vapour pressure of dichloromethane at 24.1°C is 53.3 kPa and it's enthalpy of vaporization is 28.7 kJ/mol.Estimate the temperature at which its vapour press is 70.0 kPa.[hint: P=P*e^-*]
Using the given formula, we have:
ln(P/P°) = -ΔHvap/R*((1/T) - (1/T°))
Where P° is the standard pressure (which we can take as 1 atm or 101.3 kPa), T° is the standard temperature (which we can take as 298 K), R is the gas constant (8.31 J/mol-K), ΔHvap is the enthalpy of vaporization (28.7 kJ/mol), P is the new pressure we want to find (70.0 kPa), and T is the new temperature we want to find.
Plugging in the values, we get:
ln(70.0/101.3) = -28.7/(8.31*((1/T) - (1/298)))
Simplifying this equation, we get:
ln(0.691) = -3.45*(1/T - 1/298)
ln(0.691) = -3.45/T + 3.45/298
-3.45/T = ln(0.691) - 3.45/298
T = -3.45/ln(0.691) + 3.45/(ln(0.691)*298)
T = 34.4°C
Therefore, the estimated temperature at which the vapor pressure of dichloromethane is 70.0 kPa is 34.4°C.
What do you understand by the concept of partial molar quantity in thermodynamics? Use partial molar volume as an example to explain.
In thermodynamics, a partial molar quantity represents the change in the overall property of a solution resulting from adding a small amount of a component to the solution while keeping the temperature, pressure, and other conditions constant. It is defined mathematically as the derivative of the total property with respect to the number of moles of the component added.
In other words, a partial molar quantity is the contribution of a specific component to the overall thermodynamic property of a solution, such as volume, energy, or entropy. For example, the partial molar volume of a component in a solution is the change in the volume of the solution when a small amount of the component is added, keeping temperature and pressure constant.
The partial molar volume can be calculated using the following formula:
V̅i = (∂V/∂ni)t,p
Where V̅i is the partial molar volume of component i, V is the total volume of the solution, ni is the number of moles of component i, and t and p are the constant temperature and pressure of the solution.
For instance, in a binary solution consisting of water and ethanol, the partial molar volume of ethanol can be calculated by adding a known small amount of alcohol to the solution and measuring the resulting change in volume. By applying the above formula, we can thus derive the contribution of ethanol to the overall volume of the solution, giving us insights into the thermodynamic behavior of this mixture.
The partial molar volume of water and enthanol in a solution with Xwater=0.45 at 25°C are 17.0 and 57.5 cm³/mole respectively.Calculate the volume change upon mixing sufficient ethanol with 3.75 moles of water to give this concentration. [P water =0.997gcm^-³ & P ethanol = 0.789gcm^-³ at 25 °C]
To calculate the volume change upon mixing, we can use the formula:
ΔVmix = V2 - V1
where V1 is the initial volume of the water and V2 is the final volume of the solution after adding ethanol.
We can start by calculating the initial volume of the solution, which is the volume of 3.75 moles of water:
V₁ = n₁ × M₁ / ρ₁
where n₁ is the number of moles of water, M₁ is the molar mass of water (18.015 g/mol), ρ₁ is the density of water at 25°C (0.997 g/cm³).
V₁ = 3.75 × 18.015 / 0.997
V₁ = 67.94 cm³
Next, we need to calculate the number of moles of ethanol that need to be added to the solution to obtain a mole fraction of Xethanol = 1 - Xwater = 0.55.
n₂ = (Xethanol × M₂) / (Xwater × M₁)
where M₂ is the molar mass of ethanol (46.07 g/mol).
n₂ = (0.55 × 46.07) / (0.45 × 18.015)
n₂ = 6.66 moles
The final volume of the solution can be calculated as:
V₂ = n₁V̅₁ + n₂V̅₂
where V̅₁ and V̅₂ are the partial molar volumes of water and ethanol, respectively.
V₂ = (3.75 × 17.0) + (6.66 × 57.5)
V₂ = 505.7 cm³
Therefore, the volume change upon mixing is:
ΔVmix = V₂ - V₁
ΔVmix = 505.7 - 67.94
ΔVmix = 437.8 cm³
So, the volume change upon mixing sufficient ethanol with 3.75 moles of water to give a mole fraction of 0.55 is 437.8 cm³.
The difference in chemical potential between two regions of a system is -0.82kJ/mol. By how much does the Gibbs energy change when 59 moles of substance is transferred from one region to another?
The Gibbs free energy change for a process is given by the equation:
ΔG = Δn x μ
where Δn is the change in the number of moles of substance, and μ is the chemical potential difference between the two regions of the system.
In this case, we are given that μ = -0.82 kJ/mol and that Δn = 59 mol. Therefore, we can calculate the Gibbs free energy change as:
ΔG = Δn x μ = 59 mol x (-0.82 kJ/mol)
ΔG = -48.38 kJ
The negative sign indicates that the process is exergonic, meaning that the system releases energy as substances are transferred from one region to another. So, the Gibbs energy change when 59 moles of substance is transferred from one region to another is -48.38 kJ.
The difference in chemical potential between two regions of a system is -0.82kJ/mol. By how much does the Gibbs energy change when 50 moles of substance is transferred from one region to another?
The Gibbs free energy change for a process due to the transfer of Δn moles of a substance between regions of a system, with a chemical potential difference of μ, is given by the equation:
ΔG = Δn x μ
In this case, we are given that μ = -0.82 kJ/mol and Δn = 50 mol. Therefore, we can calculate the Gibbs free energy change as:
ΔG = Δn x μ = 50 mol x (-0.82 kJ/mol)
ΔG = -41 kJ
Therefore, the Gibbs energy changes by -41 kJ when 50 moles of substance is transferred from one region to another.
For a mixture of two liquids A & B of differing volatility, their vapour pressure are PA and PB respectively.
a). Express the total pressure P in terms of partial pressure of pure A & B, and mole fraction A,is XA
b). Write an expression in terms of XA (ie make XA the subject).
c).Give an expression for component A in terms of the partial pressure of pure A&B and mole fraction A.