Question

Using the experimental value for the specific heat of gold at 155 K of cv/R=2.743, estimate the Einstein characteristic temperature (in K) - use MWAu=197.0 kg/kmol.​

Answer 1

Given that,

Temperature = 155 K    

$\dfrac{c_{v}}{R}=2.743$

We need to calculate the einstein characteristic temperature

Using formula of specific heat at low temperature

$C_{v}=3R(\dfrac{\theta_{E}}{T})^2\timese^{-\frac{\theta_{E}}{T}}$

Suppose, $\dfrac{\theta_{E}}{T}=x$

$C_{v}=3R(x^2)\timese^{-x}$...(I)

We know that,

The expansion of $e^{-x}$ is

$e^{-x}=1-\dfrac{x}{1!}+\dfrac{x^2}{2!}-\dfrac{x^3}{3!}+........</p><p>Put the value of [tex]e^{-x}$ in equation (I)

$C_{v}=3Rx^2\times(1-\dfrac{x}{1!}+\dfrac{x^2}{2!}-\dfrac{x^3}{3!}+........)$

$C_{v}=3R\times(x^2-\dfrac{x^3}{1!}+\dfrac{x^4}{2!}-\dfrac{x^5}{3!}+........)$

Now, neglecting higher powers

$C_{v}=3R\times x^2$

$C_{v}=3R\times(\dfrac{\theta_{E}}{T})^2$

$\dfrac{C_{v}}{R}=3\times(\dfrac{\theta_{E}}{T})^2$

Put the value of $\dfrac{C_{v}}{R}$ and temperature

$2.743=3\times(\dfrac{\theta_{E}}{155})^2$

$\theta_{E}^2=\dfrac{2.743\times(155)^2}{3}$

$\theta_{E}=\sqrt{\dfrac{2.743\times(155)^2}{3}}$

$\theta_{E}=148.2\ K$

Hence, The Einstein characteristic temperature is 148.2 K.

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Topic : specific heat