Calculate the change in the entropies of the system and the surroundings, and the total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 k and 1.00 bar doubles its volume in (a) an isothermal reversible expansion, (b) an isothermal irreversible expansion against pex = 0, and (c) an adiabatic reversible expansion.

Our values are:

a) For isothermal reversible expansion, We have that the change in the Internal Energy is 0, that is,

We can solve for the Work made in the system.

With work made it, this same quantity is necessary to add to the system through Heat. Then,

For Surrounding we apply the same relation,

Therefore,

b) For an isothermal irreversible expansion

We consider the external presion as 0, by this way, the work done is also 0.

Since isothermal irreversible expansion change in Internal Energy is 0, so

, since isothermal

Therefore

How the condition are the same, for the System the entropy is the same

But

c) And an adiabatic reversible expansion

Since the process is Adiabatic,

Therefore,

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz