The temperature of a sample of an ideal gas in a sealed 5.0 l container is raised from 27 oc to 77oc. if the initial pressure of the gas was 3.0 atm, what is the final pressure?

              The pressure increases to 3.5 atm.

Solution:
                According to Gay-Lussac's Law, " At constant volume and mass the pressure of gas is directly proportional to the applied temperature".

For initial and final states of a gas the equation is,

                                           P₁ / T₁  =  P₂ / T₂
Solving for P₂,
                                           P₂  =  P₁ T₂ / T₁    (1)

Data Given;
                   P₁  =  3 atm

                   T₁  =  27 °C + 273  =  300 K
 
                   T₂  =  77 °C + 273  =  350 K

Putting values in eq. 1,

                                           P₂  =  (3 atm × 350 K) ÷ 300 K

                                           P₂  =  3.5 atm

The density would be 1.622