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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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Answer:

The nail exerts a force of 573.88 Pounds on the Hammer in positive j direction.

Explanation:

Since we know that the force is the rate at which the momentum of an object changes.

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The momentum of any body is defines as \overrightarrow{p}=mass\times \overrightarrow{v}

In the above problem we see that the moumentum of the hammer is reduced to zero in 0.023 seconds thus the force on the hammer is calculated using the above relations as

\overrightarrow{F}=\frac{m(\overrightarrow{v_{f}}-\overrightarrow{v_{i}})}{\Delta t}

\overrightarrow{F}=\frac{m(0-(-73.33)}{0.23}=\frac{1.8\times 73.33}{0.23}=573.88Pounds

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Scorpion4ik [409]

Answer:

The settlement that is expected is 1.043 meters.

Explanation:

Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil

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\Delta H=\frac{H_oC_c}{1+e_o}log(\frac{\bar{\sigma_o}+\Delta \bar{\sigma }}{\bar{\sigma_o}})

where

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