Answer:- 8.97 atm.
Solution:- Moles of each gas, volume of the vessel and temperature are given. So, we could easily use the partial pressure of each gas using ideal gas law equation.
PV = nRT
P is the pressure, V is the volume, n is the moles of the gas, R is universal gas constant and T is kelvin temperature.
The equation is rearranged for the pressure as:
[tex]P=\frac{nRT}{V}[/tex]
To calculate the partial pressure of fluorine we will use its moles.
n = 2.0 mol
V = 5.0 L
T = 273 K
R = 0.0821[tex]\frac{atm.L}{mol.K}[/tex]
P = ?
Let's plug in the values and calculate the partial pressure of fluorine.
[tex]P=\frac{2.0mol*0.0821\frac{atm.L}{mol.K}*273K}{5.0L}[/tex]
P = 8.97 atm
So, the partial pressure of fluorine gas is 8.97 atm.
