Answer:
standard entropy of vaporization of ethanol = 142.105 J/K-mol
Explanation:
given data
enthalpy of vaporization of ethanol = 40.5 kJ/mol = 40.5 × 
J/mol
entropy of vaporization of ethanol boiling point = 285 K
to find out
standard entropy of vaporization of ethanol
solution
we get here standard entropy of vaporization of ethanol that is expess as
standard entropy of vaporization of ethanol ΔS = 
.............1
here ΔH is enthalpy of vaporization of ethanol and T is temperature
put value in equation 1
standard entropy of vaporization of ethanol ΔS = 
standard entropy of vaporization of ethanol = 142.105 J/K-mol
The process to create a protected area takes a long time because there's numerous things that have to be taken in account and sorted out before the area gets that kind of legal status.
Usually there's people living in the area that is supposed to be protected, though it is mostly small settlements. The people living there are connected with that area, use the resources, and their lives are largely based and dependent on it. In order for any problems to be avoided, these people should all be educated, deals to be made with them, and properly negotiated so that the area wouldn't have any damage, but also them to be able to sustain themselves.
The boundaries of the area are also complicated, it it takes time to sort out what parts should be included and what not. Basically to set the priorities.
People should be employed to protect and take care of the area, so they have to undergo a training, education, and to get familiar with it.
The legalization of the area as such can sometimes too make problems and take some time.
Answer:
a) 
b) entropy of the sistem equal to a), entropy of the universe grater than a).
Explanation:
a) The change of entropy for a reversible process:
The energy balance:
![\delta U=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=%5Cdelta%20U%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)
If the process is isothermical the U doesn't change:
![0=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=0%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)
The work:
If it is an ideal gas:
Solving:
Replacing:
Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:
b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.
