At what temperature celsius will 19.4g of molecular oxygen, o2, exert a pressure of 1820mm hg in a 5.12th cylinder?

             T  =  -28.2 °C

Solution:

Data Given;
                   P  =  1820 mmHg  =  2.39 atm

                   V  =  5.12 L

                   m  =  19.4 g

                   R  =  0.0821 atm.L.mol⁻¹.K⁻¹

Calculating moles,

                             n  = 19.4 g ÷ 32 g.mol⁻¹

                             n  =  0.606 mol

Assuming that the oxygen gas is acting perfectly, then according to Ideal Gas Equation,
                                P V  =  n R T

Solving for T,
                                T  =  P V / n R

Putting values,

              T  =  (2.39 atm × 5.12 L) ÷ (0.606 mol × 0.0821 atm.L.mol⁻¹.K⁻¹)

              T  =  244.7 K
Or,
              T  =  -28.2 °C

Spontaneity tell us the direction of the reaction, but not how fast it goes. gibbs free energy brings together enthalpy, entropy and temperature in one easy formula. g = h - ts. and, finally the sign of delta g is always negative for a spontaneous reaction.

A.