Consider a reversible isothermal expansion of a photon gas. Determine ...
The energy density of a photon gas is
U = σT4
Thus the energy is
Now use the first law of thermodynamics
dE = dQ - dW
dQ = dE + dW = dE + pdV
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Consider a reversible isothermal expansion of a photon gas. Determine ...
The energy density of a photon gas is
U = σT4
Thus the energy is
Now use the first law of thermodynamics
dE = dQ - dW
dQ = dE + dW = dE + pdV
Consider a reversible isothermal expansion of a photon gas. Determine ...
To determine the entropy S for a reversible isothermal expansion of a photon gas, we can use the formula for the entropy change during an isothermal process:
ΔS = nR ln(Vf/Vi)
Where ΔS is the change in entropy, n is the number of moles of gas, R is the gas constant, Vf is the final volume, and Vi is the initial volume.
In this case, since we are considering a photon gas, n is the number of photons and R is the specific gas constant for photons, which is given by:
R = 5/2 k
Where k is the Boltzmann constant.
Since we are assuming a reversible isothermal expansion, the initial and final temperatures will be the same, so we can write the equation as:
ΔS = (5/2) k ln(Vf/Vi)
However, since photons are massless particles, we cannot define a volume for them. Instead, we can define the volume as the volume of the container that holds the photons.
Therefore, the entropy change for a reversible isothermal expansion of a photon gas is given by:
ΔS = (5/2) k ln(Vf/Vi)
Since we do not have specific values for Vf and Vi, we cannot calculate the exact entropy change. However, we can say that the entropy will increase for an expansion (Vf > Vi), and decrease for a compression (Vf < vi).="" />