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IgorLugansk [536]
3 years ago
9

Why carbon is not used as a semiconductor material​

Engineering
1 answer:
dmitriy555 [2]3 years ago
8 0
Carbon is not used as semiconductor it has 4 valence electrons in it valence shell but the energy gap is very small it will conduct electricity even at room temperature ,the size of carbon is very small .
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Calculate the amount of current flowing through a 75-watt light bulb that is connected to a 120-volt circuit in your home.
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Consider liquid n-hexane in a 50-mm diameter graduated cylinder. Air blows across the top of the cylinder. The distance from the
ra1l [238]

The evaporation rate of the n-Hexane is 7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}

<u>Explanation</u>:

This is a situation regarding diffusing A through non-diffusing B.

A = n-Hexane B=Air

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N_{A}=D_{A B} P_{T}\left(P_{A 1}-P_{A 2}\right) / R T z P_{b m}

\mathrm{D}_{\mathrm{AB}}=8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}

P_{t}=1 a t m=101325 P a\\

\text { so, } P_{A 1}= the vapor pressure at hexane 25 \mathrm{C} =20158.2 \mathrm{Pa}

For wind, assume negligible hexane is present, hence P_{A 2}=0

Now,

\mathrm{P}_{\mathrm{B} 1}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 1}=101325-20158.2 \mathrm{P}_{\mathrm{a}}

\mathrm{P}_{\mathrm{B} 2}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 2}=\mathrm{P}_{\mathrm{T}}=101325 \mathrm{Pa}

P_{B M}=\frac{\left(P_{B 2}-P_{B 1}\right)}{\log _{e}\left(P_{B 2} / P_{B 1}\right)}\\

=\frac{101325-81166.8}{\ln \left(\frac{101325}{81166.8}\right) \mathrm{Pa}}

=90873.57 \mathrm{Pa}

R=8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K}

z=\text { distance }=20 \mathrm{cm}=0.2 \mathrm{m}\\

where T = 298 K

substituting all in the equation, we get

\begin{aligned}&\mathrm{N}_{\mathrm{A}}=\\&\left(8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}\right) \times 101325 \mathrm{Pa} \times(20158.2 \mathrm{Pa}) /(8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K} \times 0.2 \mathrm{m} \times 298 \mathrm{K}\\&\times 90873.57 \mathrm{Pa})\end{aligned}

=0.004 \mathrm{mol} / \mathrm{m}^{2} \mathrm{s}\\

Now,Flux \times area  = Molar rate of evaporation

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Evaporation rate =7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}

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Softa [21]

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