Calculate the partial pressure of 3.25 moles of oxygen that have been mixed (no reaction)
with 1.25 moles of nitrogen in a 25.0 L container of 298 K.
Use PV = nRT and substitute 3.25 for n if you want pO2.
Use PV = nRT and substitute 1.25 for n if you want pN2.
Use PV = nRT and substitute (3.25+1.25) if you want the total pressure in the container. (or add pO2 + pN2 = Ptotal).
To calculate the partial pressure of oxygen, we need to use the ideal gas law equation:
PV = nRT
where:
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature in Kelvin
First, let's calculate the partial pressure of oxygen:
Using the ideal gas law equation, we can write:
P(O2)V = n(O2)RT
Rearranging the equation to solve for P(O2):
P(O2) = (n(O2)RT) / V
Given:
n(O2) = 3.25 moles
V = 25.0 L
R = 0.0821 L·atm/mol·K
T = 298 K
Substituting the given values into the equation:
P(O2) = (3.25 * 0.0821 * 298) / 25.0
Calculating the value:
P(O2) = 7.71 atm
Therefore, the partial pressure of 3.25 moles of oxygen is 7.71 atm.