Question 16 options:
A mixture of equal moles of He and Xe are in a plastic container at 25ºC and 874 mm pressure. If only He leaks from the container and the pressure decreases to 600 mm at 25º C, what is the mole fraction of
(1) He and
(2) Xe in the container after the leak?
PV=nRT
V,R, and T remain the same before and after the leak.
Let the combined moles of He and Xe be n1 before the leak, and n2 after the leak.
Before the leak:
n1=874V/RT
Because n1 is equally split between He and Xe, the mole fraction of n1 that is He is (1/2)(874V/RT)=437V, as is the mole fraction of n1 that is Xe.
After the leak:
n2=600V/RT
n2 is not equally split because we lost some He. However, we still have the same amount of Xe, so the mole fraction of Xe is now:
(437V/RT) / (600V/RT) = 437/600
The mole fraction of He is then 1 minus this fraction.
Correction: under "Before the leak:" , 437V should be 437V/RT
To determine the mole fraction of He and Xe in the container after the leak, we need to use the ideal gas law equation and apply the concept of partial pressure.
The ideal gas law equation is given by: PV = nRT
Where:
P = pressure
V = volume
n = moles of gas
R = ideal gas constant
T = temperature in Kelvin
To find the mole fraction of He and Xe, we need to calculate the moles of He and Xe before and after the leak.
Given information:
Initial pressure (P initial) = 874 mm
Final pressure (P final) = 600 mm
Temperature (T) = 25ºC = 25 + 273 = 298 Kelvin
Step 1: Calculate the moles of He and Xe before the leak
Since the initial mixture contains equal moles of He and Xe, we can assume that the initial moles of each gas are the same.
Let's assume the initial moles of He and Xe are both 'x'.
Using the ideal gas law equation, we can rewrite it as: n = PV / RT
For He, initially: n (He initial) = P (He initial) * V / R * T
= 874 mm * V / R * 298 K
Similarly, for Xe, initially: n (Xe initial) = P (Xe initial) * V / R * T
= 874 mm * V / R * 298 K
Step 2: Calculate the moles of He after the leak
Since only He leaks from the container, the final moles of Xe will remain the same as the initial moles of Xe.
Let's assume the final moles of He are 'y'.
Using the ideal gas law equation, we can rewrite it as: n = PV / RT
For He, after the leak: n (He final) = P (He final) * V / R * T
= 600 mm * V / R * 298 K
Step 3: Calculate the mole fraction of He and Xe after the leak
To calculate the mole fraction of each gas, we need to calculate the total moles of gases after the leak.
Total moles = moles of He + moles of Xe
Therefore, Total moles = y + x
The mole fraction of He (X(He)) can be calculated using the formula: X(He) = moles of He / total moles
X(He) = y / (y + x)
Similarly, the mole fraction of Xe (X(Xe)) can be calculated as: X(Xe) = moles of Xe / total moles
X(Xe) = x / (y + x)
So, the answers to the questions are:
(1) The mole fraction of He in the container after the leak is y / (y + x).
(2) The mole fraction of Xe in the container after the leak is x / (y + x).
Please note that to get the actual numerical values, you will need to know the volume or obtain it from the given information.