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Chapter 7: Q. 7.68 (page 323)

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/cm3)

Short Answer

Expert verified

The condensate temperature for liquid helium-4 is 3.13K

Step by step solution

01

Step 1. Given information

Condensate temperature for a given substance =

Tc=1k0.527h22蟺mNV23

massofparticle,m=4(1.66脳10-27)kg

For density of H4eliquid,

mHeV=0.145g/cm3

Now,

NV=mHeVmHeN

Putting the values of mHeV=0.145g/cm3andmHeN=4.00g/mol.

NV=0.145脳103g/cm34.00g/mol

=0.036mol/cm3

=0.036mol/cm36.022脳1023atoms1mole106cm31m3

=2.18脳1028atoms/m3

02

Step 2. Putting the values of h, N/V, k, m in equation Tc=1k0.527h22πmNV23

we get,

Te=11.381脳10-23J/K(0.527)6.63脳10-34J路s22蟺(4)1.66脳10-27kg2.18脳1028atoms/m323

=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 Heatoms

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