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Natalka [10]
3 years ago
5

When the compression process is non-quasi-equilibrium, the molecules before the piston face cannot escape fast enough, forming a

high-pressure region in front of the piston. It takes more work to move the piston against this high-pressure region. Hence, a non-quasi-equilibrium compression process requires a larger work input than the corresponding quasi-equilibrium one.a. trueb. false
Engineering
2 answers:
Novay_Z [31]3 years ago
7 0

Answer:

True

Explanation:

Because in quasi static process, process occurs very slowly so it is treated as reversible process.

muminat3 years ago
5 0

Answer:

a. true

Explanation:

Firstly, we need to understand what takes places during the compression process in a quasi-equilibrium process. A quasi-equilibrium process is a process in during which the system remains very close to a state of equilibrium at all times.  When a compression process is quasi-equilibrium, the work done during the compression is returned to the surroundings during expansion, no exchange of heat, and then the system and the surroundings return to their initial states. Thus a reversible process.

While for a non-quasi equilibrium process, it takes more work to move the piston against this high-pressure region.

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Lower temperatures favor Coble creep and higher temperatures favor Nabbaro–Herring creep because the activation energy for vacancy diffusion within the lattice is typically larger than that along the grain boundaries, thus lattice diffusion slows down relative to grain boundary diffusion with decreasing temperature.
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The van der Waals equation is a modification of the ideal gas equation. What two factors does this equation account for? A. (1)
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Answer:

E

Explanation:

The vander waals equation

(P+n^2a/V^2)(V-nb)=nRT

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sertanlavr [38]

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6 0
3 years ago
The lattice constant of a simple cubic primitive cell is 5.28 Å. Determine the distancebetween the nearest parallel ( a ) (100),
skelet666 [1.2K]

Answer:

a)5.28 Å , b)3.73 Å , c)3.048 Å

Explanation:

the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.

Therefore, a particular unit cell consist only 1/8th part of an atom.

The lattice constant of a simple cubic primitive cell is 5.28 Å

We know formula of distance,

d = \frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}

a)(100)

a=5.28 Å

Distance = \frac{5.28 Å}{\sqrt{1^{2}+0^{2}+0^{2}}}=5.28 Å

b)(110)

Distance = \frac{5.28}{\sqrt{1^{2}+1^{2}+0^{2}}} = 3.73 Å

c)(111)

Distance= \frac{5.28}{\sqrt{1^{2}+1^{2}+1^{2}}}= 3.048 Å

6 0
3 years ago
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