Problem 5.4.18

Solution

Let $T_1$ denote the temperature of the outer, warmer wall of the Dewar vessel and $T_2$ the temperature of the inner wall, which is in thermal contact with the liquid air:

$$T_1=293\,\text{K},\qquad T_2=81\,\text{K}.$$

The space between the walls contains highly rarefied air. According to the condition, the distance $d$ between the walls is much smaller than the molecular mean free path $\lambda$:

$$d\ll\lambda.$$

Therefore, a molecule leaving one wall is very likely to reach the opposite wall without colliding with another molecule. Consequently, the usual continuum description of heat conduction in a gas is not applicable. Heat is transferred in the free-molecular regime, in which individual molecules carry energy directly from the warmer wall to the colder one.

We assume that after a collision with a wall, a molecule becomes thermally accommodated to that wall. Thus, molecules leaving the warm wall have a characteristic kinetic energy corresponding to the temperature $T_1$, whereas molecules leaving the cold wall have a characteristic kinetic energy corresponding to $T_2$.

The average translational kinetic energy of one molecule of an ideal gas is

$$\varepsilon=\frac32kT.$$

Hence, a molecule travelling from the warm wall to the cold wall carries an excess energy of order

$$\Delta\varepsilon=\frac32kT_1-\frac32kT_2.$$

Therefore,

$$\Delta\varepsilon=\frac32k(T_1-T_2).$$

We now estimate the number of molecules transferring energy between the walls per unit time.

Let the surface area of the walls be $S$. The volume of the rarefied gas layer between the walls is of order

$$V\sim Sd.$$

If $n$ is the molecular number density, the number of molecules contained in this volume is approximately

$$N\sim nSd.$$

The characteristic flight time of a molecule from one wall to the other is

$$t_f\sim\frac{d}{v},$$

where $v$ is a characteristic thermal speed of the molecules.

Therefore, the number of molecular crossings between the walls per unit time can be estimated as

$$\dot N\sim\frac{N}{t_f}.$$

Substituting the expressions for $N$ and $t_f$, we obtain

$$\dot N\sim\frac{nSd}{d/v}.$$

The distance between the walls cancels:

$$\dot N\sim nSv.$$

Thus, the actual value of $d$ does not enter the final estimate. Physically, increasing the separation increases the number of molecules between the walls, but it also increases the time required for each molecule to travel from one wall to the other; these two effects compensate each other.

Since the problem asks only for an estimate, numerical factors of order unity associated with the exact angular distribution of molecular velocities are neglected.

The molecular number density of the residual air follows from the ideal-gas equation:

$$p=nkT_1.$$

Hence,

$$n=\frac{p}{kT_1}.$$

As a characteristic molecular speed, we use the root-mean-square speed:

$$v=\sqrt{\frac{3kT_1}{m_0}},$$

where $m_0$ is the mass of one air molecule.

Since

$$m_0=\frac{M}{N_A}$$

and

$$R=kN_A,$$

the molecular speed can be written in terms of the molar mass $M$:

$$v=\sqrt{\frac{3RT_1}{M}}.$$

We can now determine the heat-transfer rate from the warm wall to the cold wall.

Approximately $\dot N$ molecules transfer energy per unit time, and each molecule carries an energy difference $\Delta\varepsilon$. Therefore,

$$\dot Q\sim\dot N\,\Delta\varepsilon.$$

Substituting the expressions found above gives

$$\dot Q\sim nSv\frac32k(T_1-T_2).$$

Using

$$n=\frac{p}{kT_1},$$

we obtain

$$\dot Q\sim\frac{p}{kT_1}Sv\frac32k(T_1-T_2).$$

The Boltzmann constant cancels:

$$\dot Q\sim\frac32pSv\frac{T_1-T_2}{T_1}.$$

Substituting the characteristic molecular speed gives

$$\boxed{\dot Q\sim\frac32pS\frac{T_1-T_2}{T_1}\sqrt{\frac{3RT_1}{M}}}.$$

It is useful to check the dimensions. The product $pSv$ has dimensions

$$[pSv]=\text{Pa}\cdot\text{m}^2\cdot\frac{\text{m}}{\text{s}}=\frac{\text{J}}{\text{s}}=\text{W},$$

while the temperature ratio is dimensionless. Therefore, the expression indeed has the dimensions of power.

During a time interval $\tau$, the total heat transferred to the liquid air is

$$Q=\dot Q\,\tau.$$

This heat is used to evaporate the liquid air. If a mass $m$ evaporates, then

$$Q=Lm,$$

where $L$ is the specific latent heat of vaporization.

Hence,

$$m=\frac{Q}{L}=\frac{\dot Q\,\tau}{L}.$$

Therefore,

$$\boxed{m\sim\frac{3pS\tau}{2L}\frac{T_1-T_2}{T_1}\sqrt{\frac{3RT_1}{M}}}.$$

We now substitute the numerical values:

$$p=0.133\,\text{Pa},$$

$$S=600\,\text{cm}^2=6.0\times10^{-2}\,\text{m}^2,$$

$$\tau=1\,\text{h}=3600\,\text{s},$$

$$L=0.2\,\text{MJ/kg}=2.0\times10^5\,\text{J/kg},$$

$$T_1=293\,\text{K},\qquad T_2=81\,\text{K},$$

$$M\simeq2.9\times10^{-2}\,\text{kg/mol},$$

$$R=8.31\,\text{J/(mol}\cdot\text{K)}.$$

First, the characteristic molecular speed is

$$v=\sqrt{\frac{3\cdot8.31\cdot293}{2.9\times10^{-2}}}.$$

Thus,

$$v\approx5.0\times10^2\,\text{m/s}.$$

The heat-transfer rate is therefore

$$\dot Q\sim\frac32\cdot0.133\cdot6.0\times10^{-2}\cdot5.0\times10^2\cdot\frac{293-81}{293}.$$

Hence,

$$\dot Q\approx4.3\,\text{W}.$$

During one hour, the amount of heat transferred to the liquid air is

$$Q\approx4.3\cdot3600.$$

Therefore,

$$Q\approx1.6\times10^4\,\text{J}.$$

The evaporated mass is then

$$m=\frac{1.6\times10^4}{2.0\times10^5}.$$

Thus,

$$m\approx7.8\times10^{-2}\,\text{kg}.$$

The direct estimate gives approximately $0.08\,\text{kg}$. However, the problem explicitly asks for an estimate, and the simplified free-molecular model determines the result only up to a numerical factor of order unity. Therefore, the physically appropriate order-of-magnitude answer is

$$m\sim10^{-1}\,\text{kg}.$$