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agasfer [191]
3 years ago
9

At 500oC, the diffusion coefficient of Cu in Ni is 1.3 x 10-22 m2/s, and the activation energy for diffusion is 256,000 J/mol. F

rom this information, determine the diffusion coefficient of Cu in Ni at 900 oC . (R = 8.314 J/(mol K)) 0oC = 273 K
Engineering
1 answer:
ipn [44]3 years ago
5 0

Answer:

the diffusion coefficient at 900°C is D₂=8.9*10⁻¹⁸ m²/s

Explanation:

The dependence of the diffusion coefficient is similar to the dependence of chemical reaction rate with respect to temperature , where

D=D₀*e^(-Q/RT)

where

D₀= diffusion energy at T=∞

Q= activation energy

T= absolute temperature

D= diffusion coefficient at temperature T

R=ideal gas constant

for temperatures T₁ and T₂

D₁=D₀*e^(-Q/RT₁)

D₂=D₀*e^(-Q/RT₂)

dividing both equations and rearranging terms we get

D₂=D₁*e^[-Q/R(1/T₂-1/T₁)]

replacing values

D₂=D₁*e^[-Q/R(1/T₂-1/T₁)] = 1.3*10⁻²² m²/s*e^(-256,000 J/mol/8.314 J/(mol K)*(1/1073K-1/773K) =  8.9*10⁻¹⁸ m²/s

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Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa
Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

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