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2 years ago

8

Calculate the molar heat capacity of a monatomic non-metallic solid at 500K which is characterized by an Einstein temperature of

300K. Express your result as a multiple of 3R.

1 answer:

aleksandr82 [10.1K]2 years ago

6 0

Answer:

Explanation:

Given

Temperature of solid $T=500\ K$

Einstein Temperature $T_E=300\ K$

Heat Capacity in the Einstein model is given by

$C_v=3R\left [ \frac{T_E}{T}\right ]^2\frac{e^{\frac{T_E}{T}}}{\left ( e^{\frac{T_E}{T}}-1\right )^2}$

$e^{\frac{3}{5}}=1.822$

Substitute the values

$C_v=3R\times (\frac{300}{500})^2\times (\frac{1.822}{(1.822-1)^2})$

$C_v=3R\times \frac{9}{25}\times \frac{1.822}{(0.822)^2}$

$C_v=0.97\times (3R)$

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2 years ago

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

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$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

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$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

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Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

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$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

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