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?

- 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:

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

P = ?

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

P = 8.97 atm

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

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml