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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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7. What is the voltage across a 100 ohm circuit element that draws a current of 1 A?
Marianna [84]

Answer:

100 V

Explanation:

V = IR

100 x 1 = 100

100 volts

3 0
3 years ago
An ideal Diesel cycle has a compression ratio of 17 and a cutoff ratio of 1.3. Determine the maximum temperature of the air and
Radda [10]

Answer:

maximum temperature = 1322 k

rate of heat addition = 212 kw

Explanation:

compression ratio = 17

cut off ratio = 1.3

power produced = 140 Kw

state of air at the beginning of the compression = 90 kPa and 578 c

Determine the maximum temperature of air

attached below is the detailed solution

6 0
3 years ago
Convert the unit Decimeter (dm) into Micrometer (um).
oksian1 [2.3K]

Answer:

86701 Micrometers.

Explanation:

Multiply 0.86701 dm by 100,000 to get 86701 um.

7 0
3 years ago
What is engine knock? What cause the engine knock problem?
antiseptic1488 [7]

Answer:

When the uneven burning of the fuel takes place due to the incorrect air/fuel mixture inside the engine cylinder, a knocking sound is observed. This is called as the engine knocking.

Explanation:

When the uneven burning of the fuel takes place due to the incorrect air/fuel mixture inside the engine cylinder, a knocking sound is observed. This is called as the engine knocking.

The engine knock problem can be caused due to the following reason

a) When the octane rating of the fuel used is low.

b) The deposition of the carbon around the cylinder walls takes place.

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3 0
3 years ago
If the specific surface energy for aluminum oxide is 0.90 J/m2 and its modulus of elasticity is (393 GPa), compute the critical
vampirchik [111]

Answer:

critical stress required for the propagation is 27.396615 ×10^{6} N/m²

Explanation:

given data

specific surface energy = 0.90 J/m²

modulus of elasticity E = 393 GPa = 393 ×10^{9} N/m²

internal crack length = 0.6 mm

to find out

critical stress required for the propagation

solution

we will apply here critical stress formula for propagation of internal crack

( σc ) = \sqrt{\frac{2E\gamma s}{\pi a}}    .....................1

here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm  = 0.3 ×10^{-3} m

so now put value in equation 1 we get

( σc ) = \sqrt{\frac{2E\gamma s}{\pi a}}

( σc ) = \sqrt{\frac{2*393*10^9*0.90}{\pi 0.3*10^{-3}}}

( σc ) = 27.396615 ×10^{6} N/m²

so critical stress required for the propagation is 27.396615 ×10^{6} N/m²

6 0
3 years ago
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