Question

c) What is meant by root mean square speed of a gas? Express it in terms of temperature and molecular weight of gas. Calculate rmsvfor He atoms at 300 K. (Take kg).1067.627He−×=m (1,1,3) d) What is Bose-Einstein condensation? Show that Bose-Einstein condensation temperature is given by 3/2B2V612.22π=NmkhTc (1,4) e) i) Write an expression for Planck’s law for energy density of photons in a cavity and calculate total energy density, u. ii) Consider sun as a black body whose interior consists of photons gas at K.1036×=T Calculate the energy density of the solar radiations. Take 4316kmJ107.56−−−×=σ.

Answer 1

Answer on Question #52462-Physics-Molecular Physics-Thermodynamics

c) What is meant by root mean square speed of a gas? Express it in terms of temperature and molecular weight of gas. Calculate $\mathbf{vrms}$ for $\mathbf{He}$ atoms at $300\mathbf{K}$. (Take $m_{He} = 6.67 \cdot 10^{-27} \, \text{kg}$.)

d) What is Bose-Einstein condensation? Show that Bose-Einstein condensation temperature is given by

$$T_{c} = \frac{h^{2}}{2\pi m k_{B}} \cdot \left[ \frac{N}{2.612\,V} \right]^{\frac{2}{3}}$$

e)

i) Write an expression for Planck’s law for energy density of photons in a cavity and calculate total energy density, u.

ii) Consider sun as a black body whose interior consists of photons gas at $T = 3 \cdot 10^{6} \, \text{K}$. Calculate the energy density of the solar radiations. Take $\sigma = 7.56 \cdot 10^{-16} \, \text{Jm}^{-3} \, \text{k}^{-4}$.

Solution

c) The root mean square speed is the measure of the speed of particles in a gas, defined as the square root of the average velocity-squared of the molecules in a gas. We can write it in terms of temperature and molecular weight of gas:

$$v_{rms} = \sqrt{\frac{3RT}{M_{m}}}$$

where, $R = 8.31 \frac{J}{K \cdot mol}$ is the molar gas constant, $T$ is the temperature in Kelvin and $M_{m}$ is the molar mass of the helium gas in kilograms per mole ($M_{m} = m_{He} \cdot N_{A}$, where $m_{He}$ is the mass of one molecule of the helium gas and $N_{A} = 6.022 \cdot 10^{23} \frac{1}{mol}$ is the Avogadro constant). So, for helium atoms at $300K$ we obtain:

$$v_{rms} = \sqrt{\frac{3RT}{M_{m}}} = \sqrt{\frac{3RT}{m_{He} \cdot N_{A}}} = \sqrt{\frac{3 \cdot 8.31 \frac{J}{K \cdot mol} \cdot 300K}{6.67 \cdot 10^{-27} \, \text{kg} \cdot 6.02 \cdot 10^{23} \frac{1}{\text{mol}}}} = 1365 \, \frac{\text{m}}{\text{s}}.$$

d) A Bose-Einstein condensate is a rare state (or phase) of matter in which a large percentage of bosons collapse into their lowest quantum state, allowing quantum effects to be observed on a macroscopic scale. The bosons collapse into this state in circumstances of extremely low temperature, near the value of absolute zero.

The BEC critical temperature for a uniform 3D system is given by the condition that all the particles accommodated in excited single particles state (except for a ground state) when $\mu = u_0 = 0$ are equal to the total number of particles in the system:

$$N = \int_{0}^{\infty} g(u) \overline{n_{u}} \, du.$$

where $g(u) \, du$ is the density of states in terms of the energy density:

$$g(u)du = \frac{V}{4\pi^2} \left(\frac{2m}{h^2}\right)^{\frac{3}{2}} \sqrt{u} du.$$

Thus

$$N = \frac{V}{4\pi^2} \left(\frac{2m}{h^2}\right)^{\frac{3}{2}} \int_{0}^{\infty} \sqrt{u} \frac{du}{e^{\frac{u}{k_B T}} - 1}.$$

We can evaluate the energy integral using the relation $\frac{1}{e^x - 1} = \sum_{n=1}^{\infty} e^{-nx}$, where $x = \frac{u}{k_B T}$:

$$\int_{0}^{\infty} \sqrt{u} \frac{du}{e^{\frac{u}{k_B T}} - 1} = (k_B T)^{\frac{3}{2}} \int_{0}^{\infty} \sum_{n=1}^{\infty} e^{-nx} \sqrt{x} dx = (k_B T)^{\frac{3}{2}} \sum_{n=1}^{\infty} n^{-\frac{3}{2}} \int_{0}^{\infty} e^{-t} \sqrt{t} dt = (k_B T)^{\frac{3}{2}} \zeta\left(\frac{3}{2}\right) \Gamma\left(\frac{3}{2}\right)$$

where $\zeta\left(\frac{3}{2}\right) = \sum_{n=1}^{\infty} n^{-\frac{3}{2}}$ and $\Gamma\left(\frac{3}{2}\right) = \int_{0}^{\infty} e^{-t} \sqrt{t} dt$.

Finally, we obtain the BEC critical density for a uniform 3D system:

$$n_c = \frac{N_c}{V} = \frac{1}{4\pi^2} \left(\frac{2mk_B T}{h^2}\right)^{\frac{3}{2}} \zeta\left(\frac{3}{2}\right) \Gamma\left(\frac{3}{2}\right).$$

We know that

$$\zeta\left(\frac{3}{2}\right) \cong 2.612; \Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{2}.$$

For a given density $\frac{N}{V'}$, the previous relation, written in terms of the temperature, defines the critical temperature $T_c$:

$$T_c = \frac{h^2}{2\pi m k_B} \cdot \left[ \frac{N}{2.612 V} \right]^{\frac{2}{3}}.$$

e) Planck's law for energy density of photons is

$$B_\nu(\nu, T) = \frac{2h\nu^3}{c^2} \frac{1}{e^{\frac{h\nu}{kT}} - 1}$$

where $\nu$ is frequency of photons.

Total energy density is Planck's law integrated over all frequencies:

$$P = \int_{0}^{\infty} \int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} d\nu d\theta d\phi \, B_\nu(\nu, T) \cos\theta \sin\theta = \sigma T^4.$$

Now we can find energy density of the solar radiation. Real value of Boltzmann constant is $\sigma = 5.67 \cdot 10^{-8} \frac{J}{sm^2K^4}$. Hence

$$P = \sigma T^4 = 5.67 \cdot 10^{-8} \frac{J}{sm^2K^4} (3 \cdot 10^6 K)^4 = 4.6 \cdot 10^{18} \frac{W}{m^2}.$$

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