Answer:The volume increases by 11%
Explanation:
When the pressure is decreased, the volume must definitely increase, to maintain the same pressure, the temperature must then also be increased as we see in the question. The volume was increased just as the pressure decreased and the temperature had to be increased to maintain a constant pressure. We have applied the general gas equation in solving the problem.
Given the temperature 746 K and activity of Pb equal to 0.055. The mole fraction of Pb is 0.1. So, the mole fraction of Sn = 0.9.Activity coefficient, γ = 0.055 / 0.1 = 0.55.The expression for w=ln〖γ_Pb x RT〗/(X_Sn^2 )=(-0.5978 x 8.314 J/(mol K ) x 746 K)/(0.9 x 0.9)= -4577.7 J= -4578 J
Now we use the computed value above and new temperature 773 K. The mole fraction of Sn and Pb are 0.5 and 0.5 respectively. Calculate the activity coefficient in the following manner.lnγ_Sn=w/RT X_Pb^2=(-4578 J)/(8.314 J/mol x 773 K) x 0.5 x 0.5= -0.718lnγ_Sn=exp(-0.178)=0.386The activity of Sn= γ_Sn x X_Sn=0.386 x 0.5=0.418
w of the system is -4578 J and the activity of Sn in the liquid solution of xsn at 500 degree Celsius is 0.418
H will definitely be positive because a bond is always more stable than no bond surely if it is a sigma bond.
For G you can't really know because you don't know how much energy is provided by the bond and if it outways the loss in disorder.
The reaction will become more spontaneous with a lower temperature because H tells you the reaction is exotherm
Answer:
Pieces of energy are known as Quanta.
