1 moles of a gas are initially initially under pressure 4 × 105 Pa at T=300 K. In this problem, you will want to use the followi
ng relationships:Definition of Gibbs free energy: G=U−TS+PV Equipartition: U=NDOF2NkT+constEntropy of an ideal gas: S=NklnV+const Ideal gas law: pV=NkTEquation of state for system with excluded volume b: (V−Nb)p=NkT
1)If the gas is ideal, how much does G change when the gas is allowed to expand isothermally until the pressure is reduced to 2 × 105 Pa?
Answer = -1729
2)How much did the chemical potential, ??, change in part (1)? ?final-?initial= ?
3) Now assume the gas has no interaction energy but does have an excluded volume per particle, b=1.9 × 10-28 m3, which means that S = Nk ln(V-Nb) + f(T,N) , where f is some function you won't need to know since T and N don't change. How much does G change when the gas expands isothermally until the pressure is reduced from 4 × 105 Pa to 2 × 105Pa? Gfinal-Ginitial= ?
4) How much did the chemical potential change in the process of part (3)? ?final-?initial= ?
1)If the gas is ideal, how much does G change when the gas is allowed to expand isothermally until the pressure is reduced to 2 × 105 Pa?
Answer = -1729
2)How much did the chemical potential, ??, change in part (1)? ?final-?initial= ?
3) Now assume the gas has no interaction energy but does have an excluded volume per particle, b=1.9 × 10-28 m3, which means that S = Nk ln(V-Nb) + f(T,N) , where f is some function you won't need to know since T and N don't change. How much does G change when the gas expands isothermally until the pressure is reduced from 4 × 105 Pa to 2 × 105Pa? Gfinal-Ginitial= ?
4) How much did the chemical potential change in the process of part (3)? ?final-?initial= ?
1 answer:
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Answer:
The answer is "Option A".
Explanation:
Series:
Solving the above series:
So, the series is:
Answer:
a. 130.73 atm
b. 102.62 atm
c. 87.1 atm
Explanation:
See the attached pictures.
