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lbvjy [14]
3 years ago
14

Calculate the number of vacancies per cubic meter at 1000∘C for a metal that has an energy for vacancy formation of 1.22 eV/atom

, a density of 6.25g/cm^3, and an atomic weight of 37.4 g/mol.
(A) 1.49×10^18 m^−3
(B) 7.18×10^22 m^−3
(C) 1.49×10^24 m^−3
(D) 2.57×10^24 m^−3
Engineering
1 answer:
Makovka662 [10]3 years ago
6 0

Answer:

C

Explanation:

N = Na.P/A------(1)

Na = avogadro's number = 6.02210²³

P = density

A = atomic weight of metal

When we substitute into equation 1 above we get

1.0x10²⁹atoms/m³

From here we calculate the number of vacancies

T = 1000⁰c = 1273K

The formula to use is

Nv= Nexo(Qt/K.T) -----(2)

Qt = 1.22eV

K = Boltzmann's constant = 8.6210x10^-5

When we substitute values into equation 2

We get Nv = 1.49 x 10²⁴m-3

Therefore option c is correct

Check attachment for a more detailed calculation of this question

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The best saw for cutting miter joints is the
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Travel Time Problem: Compute the time of concentration using the Velocity, Sheet Flow Method for Non-Mountainous Orange County a
Vaselesa [24]

Answer:

Total time taken = 0.769 hour

Explanation:

using the velocity method

for sheet flow ;

Tt = \frac{0.007(nl)^{0.8} }{(Pl)^{5}s^{0.4}  }  

Tt = travel time

n = manning CaH

Pl = 25years

L = how length ( ft )

s = slope

For Location ( 1 )

s = 0.045

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n = 0.06 ( from manning's coefficient table )

Tt1 = 0.128 hour

For Location ( 2 )

s = 2.5 %

L= 750

n = 0.13

Tt2 = 0.239 hour

For Location ( 3 )

s = 1.5%

L = 500 ft

n = 0.15

Tt3 = 0.237  hour

For Location (4)

s = 0.5 %

L = 250 ft

n = 0.011

Tt4 = 0.165 hour

hence the Total time taken = Tt1 + Tt2 + Tt3 + Tt4

                                              = 0.128 + 0.239 + 0.237 + 0.165 = 0.769 hour

5 0
3 years ago
Air at 80 kPa and 10°C enters an adiabatic diffuser steadily with a velocity of 150 m/s and leaves with a low velocity at a pre
il63 [147K]

Answer:

The exit temperature is 293.74 K.

Explanation:

Given that

At inlet condition(1)

P =80 KPa

V=150 m/s

T=10 C

Exit area is 5 times the inlet area

Now

A_2=5A_1

If consider that density of air is not changing from inlet to exit then by using continuity equation

A_1V_1=A_2V_2

So   A_1\times 150=5A_1V_2

V_2=30m/s

Now from first law for open system

h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w

Here Q=0 and w=0

h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}

When air is treating as ideal gas  

h=C_pT

Noe by putting the values

h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}

1.005\times 283+\dfrac{150^2}{2000}=1.005\times T_2+\dfrac{30^2}{2000}

T_2=293.74K

So the exit temperature is 293.74 K.

7 0
3 years ago
A insulated vessel s has two compartments separated by a membreane. On one side is 1kg of steam at 400 degC and 200 bar. The oth
Lilit [14]

Answer:

See explaination

Explanation:

See attachment for the detailed step by step solution of the given problem.

5 0
3 years ago
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