Investigate the chemical potential μ upon a change in volume. We will use the fact that
F=U−TS and μ=∂F∂N. We will compute μ(Vf)−μ(Vi)

Take N=20 moles (which is 20 times Avagodro's number of particles), T=300 K, Vi=0.01 m^3,Vf=0.02 m^3
For a gas with an excluded volume: U=32NkT+const S=Nkln⁡(V−bN)+const

If b=9 × 10-29 m^3, compute µ(Vf)- µ(Vi)

Explanation:

Let's rewrite F in terms of N.

C is a constant value.

We know that the chemical potential μ is the partial derivative of F with respect to N.

We can find now the chemical potential μ(Vf).

We can do the same to find μ(Vi).

Therefore µ(Vf) - µ(Vi) will be:

I hope it helps you!

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form