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Irina18 [472]
3 years ago
7

An ideal gas undergoes two processes: one frictionless and the other not. In both the cases, the gas is initially at 200 ℉ and 1

.5 atm, and when the process is over, the gas is at 700 °F and 22 atm. For which process does the gas experience a greater change in entropy?
Engineering
2 answers:
Lorico [155]3 years ago
8 0

Answer:

The change in entropy is the same in both process

Explanation:

The change is entropy is the same in both process because the entropy is about the randomness or disorderliness of the molecules of a substance due to chemical reactions or various arrangements given to them. it is a point function because it depends on the initial and final state of a substance.

In both processes the initial state of the ideal gas is the same for both process and the final state of the ideal gas for both processes are the same. hence the change in entropy is the same. also in ideal conditions there are no energy losses due to Friction.

Zarrin [17]3 years ago
6 0

Answer:

The process which has friction

Explanation:

The entropy is simply the change in the state of the things or the molecules in the system. It is simply the change in the energy of the system with a focus on the atoms in the system. This is also known as the internal energy of the system and is given the symbol, G. The friction contributes to the change in the energy of the system. This is because friction generates another form of energy - that is heat energy. This energy causes the internal temperature id the system to increase. Hence the greater change in the temperature.

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If it is desired to lay off a distance of 10,000' with a total error of no more than ± 0.30 ft. If a 100' tape is used and the
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A lake has a carrying capacity of 10,000 fish. At the current level of fishing, 2,000 fish per year are taken with the catch uni
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Answer:

The population size would be p' = 5000

The yield would be    MaxYield = 2082 \ fishes \ per \ year

Explanation:

So in this problem we are going to be examining the application of a  population dynamics a fishing pond and stock fishing and objective would be to obtain the maximum sustainable yield and and the population of the fish at the obtained maximum sustainable yield,  so basically we would be applying an engineering solution to fishing

 

    So the current  yield which is mathematically represented as

                               \frac{dN}{dt} =   \frac{2000}{1 \ year }

 Where dN is the change in the number of fish

            and dt is the change in time

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    For this we would be making use of the growth rate equation which is

                                      r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}

  Where N is the population of the fish which is given as 4,000 fishes

          and  K is the carrying capacity which is given as 10,000 fishes

             r is the growth rate

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                              r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]}  =0.833

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                      Max Yield  = \frac{rK}{4} = \frac{(10,000)(0.833)}{4} = 2082 \ fishes \ per \ year

So since the maximum sustainable yield is 2082 then the the population need to be higher than 4,000 so in order to ensure a that this maximum yield the population size denoted by p' would be

                          p' = \frac{K}{2}  = \frac{10,000}{2}  = 5000\ fishes          

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