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sleet_krkn [62]

3 years ago

5

Calculate the number of vacancies per cubic meter for some metal, M, at 749°C. The energy for vacancy formation is 0.86 eV/atom,

while the density and atomic weight for this metal are 5.94 g/cm3 (at 749°C) and 80.09 g/mol, respectively.

1 answer:

kogti [31]3 years ago

5 0

Answer:

Number of Vacancies = 1.22E23

Explanation:

Given

Energy for formation = 0.86 eV/atom (Q)

Density = 5.94 g/cm³

Atomic weight = 80.09 g/mol

Temperature = 749°C = 1022K

Avogadro's Constant = 6.022E23 atoms/mol

Boltzmann constant = 8.62E-5 eV/K (k)

Determination of the number of vacancies per cubic meter in metal, M at 749C (1022 K) is given by:

N exp (-Q/kT)

Where N = 6.022E23 * 5.94/ 80.09

N = 4.47E22

So, Number of Vacancies = 4.47E22 exp (-0.86 / (8.62E-5 * 1022))

= 4.47E22 exp (-9.76)

= 1.2150719773211E23

Number of Vacancies = 1.22E23 ------ Approximated

You might be interested in

The controller determines if a(n) _________ exists by calculating the difference between the SP and the PV.

shusha [124]

The controller determines if a(n) error exists by calculating the difference between the SP and the PV.

<h3>How does a controller work in control system?</h3>

The Control system is one where it entails if the output is one that has an effect on the input quantity.

So it uses the PV(Process Variable) set against the SP(Setpoint) to know if an error exists.

So, The controller determines if a(n) error exists by calculating the difference between the SP and the PV.

Learn more about controller from

brainly.com/question/14617664

#SPJ1

7 0

2 years ago

A satellite would have a mass of 270 kg on the surface of Mars. Determine the weight of the satellite in pounds if it is in orbi

koban [17]

Answer:

26 lbf

Explanation:

The mass of the satellite is the same regardless of where it is.

The weight however, depends on the acceleration of gravity.

The universal gravitation equation:

g = G * M / d^2

Where

G: universal gravitation constant (6.67*10^-11 m^3/(kg*s))

M: mass of the body causing the gravitational field (mass of Earth = 6*10^24 kg)

d: distance to that body

15000 miles = 24140 km

The distance is to the center of Earth.

Earth radius = 6371 km

Then:

d = 24140 + 6371 = 30511 km

g = 6.67*10^-11 * 6*10^24 / 30511000^2 = 0.43 m/s^2

Then we calculate the weight:

w = m * a

w = 270 * 0.43 = 116 N

116 N is 26 lbf

8 0

3 years ago

A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th

Taya2010 [7]

Answer:

(a) The heat transferred is 2552.64 kJ

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588 = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 = 2552.64 kJ

b) $\Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)$

Where:

$x_A$ and $x_B$ are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, $x_A$ = 248 /(248 + 42.8) = 0.83

$x_B$ = 42.8/(248 + 42.8) = 0.1472

∴ $\Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472)$ = 1066.0279 J/K

5 0

3 years ago

A bus travels the 100 miles between A and B at 50 mi/h and then another 100 miles between B and C at 70 mi/h.

stira [4]

Answer:

c. less than 60 mi/h

Explanation:

To calculate the average speed of the bus, we need to calculate the total distance traveled by the bus, as well as the total time of travel of the bus.

Total Distance Traveled = S = 100 mi + 100 mi

S = 200 mi

Now, for total time, we calculate the times for both speeds from A to b and then B to C, separately and add them.

Total Time = t = Time from A to B + Time from B to C

t = (100 mi)/(50 mi/h) + (100 mi)(70 mi/h)

t = 2 h + 1.43 h

t = 3.43 h

Now, the average speed of bus will be given as:

Average Speed = V = S/t

V = 200 mi/3.43 h

<u>V = 58.33 mi/h</u>

It is clear from this answer that the correct option is:

<u>c. less than 60 mi/h</u>

7 0

3 years ago

Tensile Strength (MPa) Number-Average Molecular Weight (g/mol)

IceJOKER [234]

Answer:

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

$\mathbf{M_n = 49163.56431 \ g/mol }$

Explanation:

The question can be well structured in a table format as illustrated below:

Tensile Strength (MPa) Number- Average Molecular Weight (g/mol)

82 12,700

156 28,500

The tensile strength and number-average molecular weight for two polyethylene materials given above.

Estimate the number-average molecular weight that is required to give a tensile strength required above. Using the data given find TS (infinity) in MPa.

<u>SOLUTION:</u>

We know that :

$T_S = T_{S \infty} - \dfrac{A}{M_n}$

where;

$T_S$ = Tensile Strength

$T_{S \infty}$ = Tensile Strength (Infinity)

$M_n$ = Number- Average Molecular Weight (g/mol)

SO;

$82= T_{S \infty} - \dfrac{A}{12700} ---- (1)$

$156= T_{S \infty} - \dfrac{A}{28500} ---- (2)$

From equation (1) ; collecting the like terms; we have :

$T_{S \infty} =82+ \dfrac{A}{12700}$

From equation (2) ; we have:

$T_{S \infty} =156+ \dfrac{A}{28500}$

So; $T_{S \infty} = T_{S \infty}$

Then;

$T_{S \infty} =82+ \dfrac{A}{12700} =156+ \dfrac{A}{28500}$

Solving by L.C.M

$\dfrac{82(12700) + A}{12700} =\dfrac{156(28500) + A}{28500}$

$\dfrac{1041400 + A}{12700} =\dfrac{4446000 + A}{28500}$

By cross multiplying ; we have:

$({4446000 + A})* {12700} ={28500} *({1041400 + A})$

$(5.64642*10^{10} + 12700A) =(2.96799*10^{10}+ 28500A)$

Collecting like terms ; we have

$(5.64642*10^{10} - 2.96799*10^{10} ) =( 28500A- 12700A)$

$2.67843*10^{10} = 15800 \ A$

Dividing both sides by 15800:

$\dfrac{ 2.67843*10^{10} }{15800} =\dfrac{15800 \ A}{15800}$

A = 1695208.861

From equation (1);

$82= T_{S \infty} - \dfrac{A}{12700} ---- (1)$

Replacing A = 1695208.861 in the above equation; we have:

$82= T_{S \infty} - \dfrac{1695208.861}{12700}$

$T_{S \infty}= 82 + \dfrac{1695208.861}{12700}$

$T_{S \infty}= \dfrac{82(12700) +1695208.861 }{12700}$

$T_{S \infty}= \dfrac{1041400 +1695208.861 }{12700}$

$T_{S \infty}= \dfrac{2736608.861 }{12700}$

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

From equation(2);

$156= T_{S \infty} - \dfrac{A}{28500} ---- (2)$

Replacing A = 1695208.861 in the above equation; we have:

$156= T_{S \infty} - \dfrac{1695208.861}{28500}$

$T_{S \infty}= 156 + \dfrac{1695208.861}{28500}$

$T_{S \infty}= \dfrac{156(28500) +1695208.861 }{28500}$

$T_{S \infty}= \dfrac{4446000 +1695208.861 }{28500}$

$T_{S \infty}= \dfrac{6141208.861}{28500}$

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

We are to also estimate the number- average molecular weight that is required to give a tensile strength required above.

If the Tensile Strength (MPa) is 82 MPa

Definitely the average molecular weight will be = 12,700 g/mol

If the Tensile Strength (MPa) is 156 MPa

Definitely the average molecular weight will be = 28,500 g/mol

But;

Let us assume that the Tensile Strength (MPa) = 181 MPa for example.

Using the same formula:

$T_S = T_{S \infty} - \dfrac{A}{M_n}$

Then:

$181 = 215.481- \dfrac{1695208.861 }{M_n}$

Collecting like terms ; we have:

$\dfrac{1695208.861 }{M_n} = 215.481- 181$

$\dfrac{1695208.861 }{M_n} =34.481$

1695208.861= 34.481 $M_n$

Dividing both sides by 34.481; we have:

$M_n$ = $\dfrac{1695208.861}{34.481}$

$\mathbf{M_n = 49163.56431 \ g/mol }$

5 0

3 years ago