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hodyreva [135]
3 years ago
10

Given one mole of diamond vs one mole of graphite,

Physics
2 answers:
grandymaker [24]3 years ago
7 0

Answer:

The pressure is P= -  6.39*10^8Pa

The temperature is T =1218.63 K

Explanation:

Generally Gibbs free energy is mathematically represented as

                   G = E + PV -TS

   Where  E is the enthalpy

               PV is the pressure volume energy (i.e PV energy)

                S  is the entropy

                T is the temperature

For stability to occur the Gibbs free energy must be equal to zero

Considering Diamond

  So at temperature of  T = 300 K

         E + PV - TS = 0

making P the subject

          P = \frac{TS-E}{V}

Now substituting 300 K for T , 2900 J  for E ,

                              3.42cm^3 = \frac{3.42}{1*10^6} = 3.42*10^{-6}m^3 for V and 2.38 J/K for S

     P = \frac{(300 * 2.38)- 2900}{3.42*10^{-6}}

         P= -  6.39*10^8Pa

The negative sign signifies the direction of the pressure

Given that  P = 1*0^5Pa

making T the subject

            T = \frac{PV+E}{S}

Substituting into the equation

            T = \frac{1*10^5 * 3.42 *10^{-6}+2900}{2.38}

                T =1218.63 K

             

         

kvv77 [185]3 years ago
6 0

Correct Data:

                                      G                       S                 V   

1  mol Graphite             0 J               5.74 J/K     5.30cm³

1 mol Diamond           2900 J          2.38 J/K     3.42cm³

Answer:

a) P = -639 MPa

b) T =1218.63 K

Explanation:

a) For stability of diamond, the gibb's free energy, G = 0 J

The Gibb's free energy is given by G = E + PV - TS

E + PV - TS = 0

where, E = enthalpy, P = pressure, S = entropy

E = 2900 J

T = 300 K

S = 2.38 J/K

V = 3.42 cm³ = 3.42 * 10⁻⁶ m³

2900 + P3.42 * 10⁻⁶  - 300(2.38) = 0

P3.42 * 10⁻⁶  = -2186

P = 2186/(3.42 * 10⁻⁶)

P = -639* 10⁶ Pa

P = -639 MPa

b) At P = 10⁵

E + PV - TS = 0

2900 + 10⁵(3.42 * 10⁻⁶) - T(2.38) = 0

T =1218.63 K

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Explanation:

The surface of the ramp here is assumed to be the positive x-direction.

To solve this problem and find the magnitude of the force of tension, we have to analyze only the situation along the x-direction, since the force of tension lie in this direction.

There are three forces acting along the x-direction:

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We know that the magnitude of the weight is

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So its x-component is

F_{gx}=F_g sin \theta =(70.0)(sin 22^{\circ})=26.2 N

The net force along the x-direction can be written as

F_x = F_T-F_f-F_{gx}

And therefore, since the net force is 98 N, we can find the magnitude of the force of tension:

F_T=F_x+F_f+F_{gx}=98+14.8+26.2=139 N

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3 years ago
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A car accelerates from 20mi/hr to 60mi/hr. How many times greater is the car's kinetic energy at the higher speed compared to th
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Ask Your Teacher Suppose the roller coaster below(h1 = 36 m, h2 = 13 m, h3 = 30) passes point A with a speed of 1.00 m/s. If the
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Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = (\frac{1}{5} × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

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3 years ago
A homeowner uses a snowblower to clear his driveway. knowing that the snow is discharged at an average angle of 40° with the hor
Nesterboy [21]

given that snow is projected at an angle of 40 degree

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R = \frac{v_o^2 sin2\theta}{g}

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Veseljchak [2.6K]

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Explanation:

maximum wavelength, λ = 542 nm = 542 x 10^-9 m

The work function of the metal is defined as the minimum amount of energy falling on the metal so that the photo electrons just ejects the surface of metal.

W_{o}=\frac{hc}{\lambda }

where, h is the Plank's constant and c be the speed of light

h = 6.634 x 10^-34 Js

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W_{o}=\frac{6.634\times 10^{-34}\times 3\times 10^{8}}{542\times10^{-9} }

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Wo = 2.295 eV

Thus, the work function of this metal is 2.295 eV.

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