Air is basically 80% of nitrogen and 20% of oxygen. a 2.0 mol sample of air is found to occupy 6.0l at 27%. what is the partial pressure of oxygen in the sample?

Explanation:

1. Moles of O₂  

2. Partial pressure of O₂

To calculate the partial pressure of the oxygen, we can use the Ideal Gas Law:

pV = nRT

Data:

V = 6.0 L

n = 0.40 mol

T = 27 °C

Calculations:

T = (27 + 273.15) K = 300.15 K

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