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Homework Statement
A small sphere, with initial temperature ##T##, is immersed in an ideal Boltzmannian gas at
temperature ##T_0##. Assuming that the molecules incident on the sphere are first absorbed and then re-emitted with the temperature of the sphere, determine the variation of the temperature of the sphere with time.
[Note: The radius of the sphere may be assumed to be much smaller than the mean free path of the molecules.]
This is problem 6.17 from Pathria.
The Attempt at a Solution
We first find the number of particles which strike the sphere per unit area per unit time i.e. ##\frac{dN}{dA dt}##. First consider a surface element ##dA## of the sphere. Let ##f(v)## be the thermalized Maxwell-Boltzmann distribution of the gas. We choose a local spherical coordinate system at ##dA## with the ##z## axis aligned with the normal to ##dA##. Then the number of gas particles with velocity between ##\vec{v}## and ##\vec{v} + d\vec{v}## which strike ##dA## in a time ##dt## is ##f(v)v d^3 \vec{v} dt dA \cos\theta## where ##\theta## is the angle ##\vec{v}## makes with ##dA## i.e. the local ##z## axis.
Then ##\frac{dN}{dA dt} = \int f(v) d^3 \vec{v} \cos\theta = 2\pi \int_0 ^{\infty} v^3 f(v) dv \int _{0}^{\pi/2}\sin\theta \cos\theta d\theta## in the local spherical coordinates. Here ##\theta## only goes from ##[0, \pi/2]## instead of the usual ##[0,\pi]## because the range of velocities corresponding to ##[\pi/2,\pi]## represents gas particles moving away from the local ##z## axis so they won't be colliding with the sphere. This then yields ##\frac{dN}{dA dt} = \frac{1}{4}n \langle v \rangle## where ##n \equiv \frac{N}{V}## is the number density, which is uniform for a thermalized ideal gas, and ##\langle v \rangle = \frac{4\pi}{n} \int _0 ^{\infty} v^3 f(v) dv## is the average speed of each particle. Note that for a thermalized ideal gas ##\langle v \rangle## is isotropic (an explicit expression can be obtained from the Maxwell-Boltzmann distribution ##f(v)## which I will write down at the very end below). Thus by integrating over the whole sphere we get ##\frac{dN}{dt} = \frac{1}{4}n \langle v \rangle \int dA = \pi R^2 n \langle v \rangle ## where ##R## is the sphere radius; this is the number of particles which strike the entire sphere per unit time.
Assume that the gas particles incident on the sphere are re-emitted instantaneously after being absorbed. If the gas is monoatomic then the average energy per particle before impinging on the sphere is given by ##\langle \epsilon \rangle = \frac{3}{2}k_B T_0## which comes from its heat capacity ##^{\dagger}##. The assumption here is that the gas particles thermalize after being re-emitted by the sphere before colliding again with the sphere. This is valid since the radius of the sphere is much smaller than the mean free path of the gas. The change in average energy per particle after being absorbed and remitted is ##d \langle \epsilon \rangle = \frac{3}{2}k_B [T(t) - T_0]## where ##T(t)## is the temperature of the sphere at that instant.
By conservation of energy, the energy of the sphere changes by the amount ##d\epsilon = - \frac{3}{2}k_B [T(t) - T_0]##.
The rate of change of the total energy ##E## of the sphere is then ##\frac{dE}{dt} = d\epsilon \frac{dN}{dt} = - \frac{3}{2}\pi R^2 k_B n \langle v \rangle [T(t) - T_0]##.
Now ##\frac{dE}{dt} = \frac{\partial E}{\partial T})_V \frac{dT}{dt} = C_V \frac{dT}{dt}## where ##C_V## is the constant volume heat capacity of the sphere so we have ##\frac{dT}{dt} = - \frac{\frac{3}{2}\pi R^2 k_B n \langle v \rangle}{C_V} [T(t) - T_0]##. Note ##C_V## is clearly independent of the time ##t## since it's an intrinsic property of the sphere and we are assuming there are no changes to the intrinsic properties of the sphere; furthermore ##n, \langle v \rangle## are also time-independent since the ideal gas is thermalized. Therefore we can straightforwardly integrate ##\frac{dT}{dt}## using ##T(0) = T## to get ##T(t) = T_0 + (T - T_0)e^{-\frac{\frac{3}{2}\pi R^2 k_B n \langle v \rangle}{C_V} t}## as the temperature variation ##^{\dagger \dagger}##.
##^{\dagger}## If the gas is not monoatomic, the equipartition theorem for the average energy per particle would still hold except it would be ##\langle \epsilon \rangle = \frac{\kappa}{2}k_B T## where e.g. ##\kappa = 5## if the particle has internal rotational degrees of freedom or ##\kappa = 7## if it has on top of this internal vibrational degrees of freedom. So I don't think the calculation would change in any way, is this correct?
The more important question is, I used the (equipartition) average energy per particle when considering conservation of energy for the sphere. Is this valid? Is it ok to consider the average energy per particle for conservation of energy instead of the purely microscopic Hamiltonian of the particles? I think it is because I see no other way but to use the heat capacity ##\frac{\partial \langle \epsilon \rangle}{\partial T})_V## to find the change in energy of the gas particles due to the change in temperature upon being re-emitted by the sphere. Temperature would not even enter into a purely microscopic Hamiltonian consideration.
##^{\dagger \dagger}## Is this answer actually correct? My issue is the problem doesn't provide the radius ##R## of the sphere or its heat capacity ##C_V## and I can't see any possible way to re-express ##R## or ##C_V## in terms of the parameters given in the problem. My expression for ##T(t)## does have the right limiting cases though. The characteristic time of ##\frac{dT}{dt}## is ##\tau = \frac{C_V}{\frac{3}{2}\pi R^2 k_B n \langle v \rangle}##. As ##C_V \rightarrow \infty##, ##\tau \rightarrow \infty## as we would expect and as ##R \rightarrow \infty##, ##\tau \rightarrow 0## which is also to be expected since there would be more collisions of gas particles with the sphere. And of course ##T(t) \rightarrow T_0## as ##t \rightarrow \infty## i.e. the sphere comes to equilibrium with the gas.
Thanks in advance!
As an aside let me just note that from the explicit expression for the Maxwell-Boltzmann distribution ##f(v)## it can be shown that ##\frac{1}{4}n \langle v \rangle = \frac{P}{\sqrt{2\pi m k_B T}}## where ##P## is the pressure of the gas.
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