Question

5.0 L vessel holds 3.0 mol n2, 2.0 mol f2, and 1.0 mol h2 at 273 K. What is the partial pressure of fluorine?

Answer 1

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:

`P=(nRT)/(V)`

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
`(atm.L)/(mol.K)`

P = ?

Let's plug in the values and calculate the partial pressure of fluorine.

`P=(2.0mol*0.0821(atm.L)/(mol.K)*273K)/(5.0L)`

P = 8.97 atm

So, the partial pressure of fluorine gas is 8.97 atm.