4
$\begingroup$

In Physical Chemistry we are working a problem that is confusing. We started by calculating the pressure half way to the centre of the sun assuming that the interior consisted of ionized hydrogen atoms. We determined this to be $\pu{3.59e6 kPa}$. However, the third part of the question asks:

What is the kinetic-energy density half way to the centre of the Sun? Compare with the (translational) kinetic energy density of Earth's atmosphere.

Can someone expand on what kinetic energy density is, if it is different from kinetic energy, and what the formula is. I might just be overthinking the problem.

$\endgroup$
  • 2
    $\begingroup$ I would guess that this is asking you to use the Maxwell-Boltzmann distribution, but with the kinetic energy substitution. Namely, make the substitution $v=\sqrt{\frac{2E}{m}}$ and then take the derivative to substitute $dE$ for $dv$. So, that gives you the energy distribution and then the energy density is just this divided by the pressure you determined (maybe multiplied by a constant). Notice that [N]/[m^2] is dimensionally the same as [J]/[m^3]. Probably the simpler approach is to use a formula for average velocity then get KE that way. $\endgroup$ – jheindel Aug 25 '17 at 3:53
  • 1
    $\begingroup$ I recognize that this is a crude approximation to use all these ideal gas equations, but without more information I would guess this is what you're supposed to do. $\endgroup$ – jheindel Aug 25 '17 at 3:54
  • $\begingroup$ I'm voting to close this question as off-topic because it is dealing with physics, not chemistry. $\endgroup$ – Jan Aug 25 '17 at 12:49
  • 2
    $\begingroup$ This is a question from Atkins' Physical Chemistry. This question deals equally with physics and chemistry and I'm against closing it. The textbook defines kinetic energy density as $\rho_\text{k}=E_\text{k}/V$, and in combination of kinetic energies of the particles $E_\text{k}=1/2NMc^2$ with the mass density $\rho_\text{k}=3/2p$ ($p$ -- pressure in the interior of the Sun). The textbook also gives $\rho_\text{k}(1/2r_\odot)=\pu{0.11 GJ cm-3}$, $\rho_\text{k}(\text{atm})=\pu{0.15 J cm-3}$. $\endgroup$ – andselisk Aug 25 '17 at 13:31
  • 1
    $\begingroup$ OP also missed as essential part of the question (Q1.39c from Elements of physical chemistry): "[...] Compare your result with the (translational) kinetic energy density of the Earth’s atmosphere on a warm day (25°C): $\pu{1.5e5 J m−3}$ (corresponding to $\pu{0.15 J cm−3}$)." Probably you want to add this into the question because it is rather important. $\endgroup$ – andselisk Aug 25 '17 at 13:44
4
$\begingroup$

I would like to first point out that the pressure you've determined is too low, most likely because you used only molar mass of protons $\ce{H+}$ instead of average mass for star matter in order to fulfill the condition of plasma quasineutrality:

$$\bar{M} = \frac{M(\ce{H+}) + M(\ce{e-})}{2} = \frac{\pu{(1 + 0) g mol-1}}{2} = \pu{0.5 g mol-1} = \pu{5e-4 kg mol-1} \tag{1.1}$$

Referring to the question from Atkins' Elements of Physical Chemistry [1, p. 40], where this question has been taken from, we obtain necessary data to complete the solution:

In the Sun [...] the density is 150 times that of liquid water at its centre and comparable to that of water about half-way to its surface

and

(a) Calculate the pressure half-way to the centre of the Sun, assuming that the interior consists of ionized hydrogen atoms, the temperature is $\pu{3.6 MK}$, and the mass density is $\pu{1.20 g cm−3}$ (slightly higher than the density of water)

the pressure of star plasma (that can be treated as an ideal gas) can be found as following:

$$p = \frac{\rho RT}{\bar{M}} = \frac{\pu{1.2e3 kg m-3} \cdot \pu{8.314 J K-1 mol-1} \cdot \pu{3.6e6 K}}{\pu{5e-4 kg mol-1}} = \pu{7.18e13 Pa} \label{eqn:1.2}\tag{1.2}$$


Now for the second part, kinetic energy density is defined as kinetic energy per volume:

$$\rho_\text{k} = \frac{E_\text{k}}{V} \label{eqn:2.1}\tag{2.1}$$

Kinetic energy for a set of entities (ion, atom, molecule) is

$$E_\text{k} = \frac{Nm\bar{v^2}}{2} \label{eqn:2.2}\tag{2.2}$$

where $N$ is the number of particles, $m$ is the particle's mass, $\bar{v}$ is an average (root mean square) particle velocity. From the kinetic model of an ideal gas, the pressure obtained from \eqref{eqn:1.2} is also

$$p = \frac{Nm\bar{v^2}}{3V} \label{eqn:2.3}\tag{2.3}$$

Equation \eqref{eqn:2.3} implies $Nm\bar{v^2} = 3pV$, so kinetic energy from \eqref{eqn:2.2} can be rewritten as $E_\text{k} = \displaystyle\frac{3pV}{2}$. Finally, kinetic energy density, taking \eqref{eqn:2.1} into account, is

$$\rho_\text{k} = \frac{3p}{2} = \frac{3 \cdot \pu{7.18e13 Pa}}{2} = \pu{1.08e14 J m-3} \quad (\pu{1 J} = \pu{1 Pa}\cdot \pu{1 m3})$$

Comparison between kinetic energy densities $\rho_\text{k}(1/2r_\odot)$ (halfway to the Sun's center) and $\rho_\text{k}(\text{atm})$ (Earth's atmosphere) is really a matter of comparing the pressures:

$$\frac{\rho_\text{k}(1/2r_\odot)}{\rho_\text{k}(\text{atm})} = \frac{p(1/2r_\odot)}{p(\text{atm})} = \frac{\pu{7.18e13 Pa}}{\pu{101325 Pa}} = 7.09\cdot10^8$$

References

  1. Atkins, P. W.; De Paula, J. Elements of physical chemistry, 5th ed.; Oxford University Press: Oxford; New York, 2009. ISBN 978-1-4292-1813-9.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.