University:
The Pennsylvania State University
Course:
CHEM 450 | Thermodynamics
Academic year:

2024
Views:

190

Pages:

2
Author:

Alexis H.

Problem Set #34 Assigned: March 27, 2024 Due: March 29, 2024 at 2:30 pm 1. Using appropriate equations, explain how Raoult’s and Henry’s law can be used to specify the chemical potential for component A of a two component mixture. 2. Explain the molecular level origins of Raoult’s and Henry’s law. Use appropriate figures to justify your answer. 3. The vapor pressure of pure benzene at 20 °C is 0.100 bar, while the vapor pressure of pure methylbenzene is 0.028 bar. What is the vapor pressure of a solution containing an equal mass of benzene and methylbenzene? The molecular weight of benzene is 78.11 g/mol while the molecular weight of methylbenzene is 94.12 g/mol. You may assume that benzene and methylbenzene form an ideal solution. 4. The Henry’s law constant for CO2 in water at 298 K is: KB = 30.1 bar Kg/mol (a) Find the solubility of CO2(g) in water in units of molality when its partial pressure is 0.100 bar. (b) Next, find the molality when the partial pressure is increased to 1.00 bar. 1. Raoults law relates to the vapor pressure of an ideal solution, which assumes that the components in the mixture behave ideally. According to Raoults, the partial pressure of a component in an ideal mixture is directly proportional to its mole fraction in the mixture M Mo RT In Henry’s law relates to the solubility of a component in a solvent, particularly gases dissolved in liquids. According to Henry’s law, the concentration of a gas component in a solution is directly proportional to its partial pressure above the solution µ Not RTIn 2. The molecular origin of Raoult’s law is based on the idea that the interactions between di erent molecules are similar to the interactions between like molecules. The molecular origin of Henry’s law is the gas-liquid interactions, the rate at which gas molecules enter the liquid is proportional to the gas’s partial pressure; higher pressure means more molecules are pressing down on the liquid, increasing the frequency of collisions and the dissolution rate 9 3. my 0.028bar 92.14 mol 0.00128 not 0.0003 n notNMB n 0.00128 0.0003 0.00128 B 0 810 000158 MB 809 P B Pi P XmPii 0.190 0.810 0.100 0.1900.028 0.00532bar P PBPm 0081 0.00532 L a 0.081bar 0.08632 S KP s 3015949 0.100bar s 3.01 do b S KP S 30.1barn S 30 I mk8 m 36.1 m 0033 7 Ibar