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, $2.17 K$ . ( the density of liquid helium-4 is $0.145 g / c m^{3}$ )

Short Answer

Expert verified

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

Step by step solution

Step 1. Given information

Condensate temperature for a given substance =

$T_{c} = \frac{1}{k} 0.527 (\frac{h^{2}}{2 π m}) {(\frac{N}{V})}^{\frac{2}{3}}$

$m a s s o f p a r t i c l e, m = 4 (1.66 \times 10^{- 27}) k g$

For density of ${}^{4}H e$ liquid,

$\frac{m_{H e}}{V} = 0.145 g / {cm}^{3}$

Now,

$\frac{N}{V} = [\frac{(\frac{m_{H e}}{V})}{(\frac{m_{H e}}{N})}]$

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

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

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

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

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

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

we get,

$T_{e} = \frac{1}{(1.381 \times 10^{- 23} J / K)} (0.527) \frac{{(6.63 \times 10^{- 34} J \cdot s)}^{2}}{(2 π (4) (1.66 \times 10^{- 27} kg))} {(2.18 \times 10^{28} atoms / m^{3})}^{\frac{2}{3}}$

$= 3.13 K$

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 $H e$ atoms

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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, $2.17 K$ . ( the density of liquid helium-4 is $0.145 g / c m^{3}$ )

Short Answer

Step by step solution

Step 1. Given information

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

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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

Step by step solution

Step 1. Given information

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Rotational Dynamics

Energy Physics

Nuclear Physics

Fields in Physics

Torque and Rotational Motion

Modern Physics

Study anywhere. Anytime. Across all devices.

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, $2.17 K$ . ( the density of liquid helium-4 is $0.145 g / c m^{3}$ )