- solve what volume is occupied by 5.024g of O2 at 28.0 C and a pressure of .998 atm? -

V = 3.89 L

Explanation:

Given data:

Mass of oxygen  = 5.024 g

Temperature = 28°C (28+273 = 301 k)

Pressure = 0.988 atm

Volume occupy by oxygen = ?

Solution:

PV = nRT

We will calculate the number of moles;

Number of moles = mass/ molar mass

Number of moles = 5.024 g/ 32 g/mol

Number of moles = 0.157 mol

Now we will calculate the volume,

PV = nRT

V = nRT/P

V = 0.157 mol × 0,.0821 atm.L /mol.K × 301 K / 0.998 atm

V = 3.88 L / 0.998

V = 3.89 L

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