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 $0.145g/c{m}^3$)

Short answer

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

Step-by-step solution

Step 1. Given information

Condensate temperature for a given substance =

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

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

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 Tc=1k0.527h22πmNV23

we get,

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

$=3.13K$

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

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