Q. 7.68 - An Introduction to Thermal Physics

Question

Calculate the condensate temperature for liquid helium-4, pretending that liquid is a gas of noninteracting atoms. Compare to the observed  temperature of the superfluid transition, $2.17K$. ( the density of liquid helium-4 is $0.145 g/c{m}^3$)

Step-by-Step Solution

Answer

The condensate temperature for liquid helium-4 is $3.13K$

Step 1. Given information

Condensate temperature for a given substance =

$T_c=\frac{1}{k}0.527\left(\frac{h^2}{2\pi m}\right){\left(\frac{N}{V}\right)}^{\frac{2}{3}}$

$mass of particle, m = 4(1.66\times {10}^{-27}) kg$

For density of ${}^{4}He$ liquid,

$\frac{m_{He}}{V}=0.145 \mathrm{g}/{\mathrm{cm}}^3$

Now,

$\frac{N}{V}=\left[\frac{\left(\frac{m_{He}}{V}\right)}{\left(\frac{m_{He}}{N}\right)}\right]$

Putting the values of $\frac{m_{\text{He }}}{V}=0.145 \mathrm{g}/{\mathrm{cm}}^3\text{ and }\frac{m_{\mathrm{He}}}{N}\text{= }4.00 \mathrm{g}/\mathrm{mol}\text{ .}$

$\frac{N}{V}=\left(\frac{0.145\times {10}^3 \mathrm{g}/{\mathrm{cm}}^3}{4.00 \mathrm{g}/\mathrm{mol}}\right)$

     $=0.036 \mathrm{mol}/{\mathrm{cm}}^3$

     $=0.036 \mathrm{mol}/{\mathrm{cm}}^3\left(\frac{6.022\times {10}^{23}\text{ atoms }}{1 \mathrm{mole}}\right)\left(\frac{{10}^6{ \mathrm{cm}}^3}{1{ \mathrm{m}}^3}\right)$

      $=2.18\times {10}^{28}\text{ atoms }/{\mathrm{m}}^3$

Step 2. Putting the values of h ,   N / V ,   k ,   m in equation T c = 1 k 0 . 527 h 2 2 π m N V 2 3

we get,

$T_e=\frac{1}{\left(1.381\times {10}^{-23} \mathrm{J}/\mathrm{K}\right)}(0.527)\frac{{\left(6.63\times {10}^{-34} \mathrm{J}·\mathrm{s}\right)}^2}{\left(2\pi \left(4\right)\left(1.66\times {10}^{-27} \mathrm{kg}\right)\right)}{\left(2.18\times {10}^{28}\text{ atoms }/{\mathrm{m}}^3\right)}^{\frac{2}{3}}$

      $=3.13 K$

 

Therefore the condensate temperature for liquid helium-4 is $3.13 K$.

This value is similar to experimental value of $2.17 K$, neglecting all the interaction between $He$ atoms