Answered: ƏFint ƏF S= - 51 + NvQ V = Nk (6.92) ƏT…

Related questions

Question

Equations 6.92 and 6.93 (See attached) for the entropy and chemical potential involve the logarithm of the quantity V Zint!N vQ. Is this logarithm normally positive or negative? Plug in some numbers for an ordinary gas and discuss.

ƏFint
ƏF
S= -
51
+
NvQ
V
= Nk
(6.92)
ƏT
ƏT
V,N
ƏF
(V Zint
(6.93)
u =
= -kT In
ON
/T,V
NvQ

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Similar questions

Please help me
Take V2 final volume to be 3 liters.
A certain Carnot engine operates between 290 K and 450 K. The engine does 150 J of work for every 600 J of heat absorbed from the combustion of fuel. Find the actual efficiency and the theoretical efficiency. Find the heat exhausted to the cold reservoir. Find the entropy change of the hot reservoir and the cold reservoir, and the net change of entropy of the entire engine. (For this finding the change in entropy, Qh is negative as energy is leaving the reservoir, Qc is positive)
Answer in 90 minutes please.
Polymers, like rubber, are made of very long molecules, usually tangled up in a configuration that has lots of entropy. As a very crude model of a rubber band, consider a chain of N links, each of length L Imagine that each link has only two possible states, pointing either left or right. The total length L of the rubber band is the net displacement from the beginning of the first link to the end of the last link. Using the thermodynamic identity, you can now express the tension force F in terms of a partial derivative of the entropy. From this expression, compute the tension in terms of L, T , N, and l.
Suppose that 10 distinguishable particles are equipartitioned in a container that has 100 equal-sized compartments. What is the entropy on this system?

ƏFint ƏF S= - 51 + NvQ V = Nk (6.92) ƏT ƏT V,N ƏF (V Zint (6.93) u = = -kT In ON /T,V NvQ