# What would be pressure of 1 kg of photon gas at room temperature put in a volume of 1 liter?

Suppose a number of photons with spectrum corresponding to black body spectrum at 293 K with total energy corresponding to 1 kg put in a box with ideal mirror walls with volume of 1/1000 of a cubic meter (1 liter). What pressure this photonic gas will manifest on the walls of the box?

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Homework? If so, should be marked as such. – mbq Dec 31 '11 at 16:58
It certainly looks like homework... Anixx, if you didn't have such a well-established history here of posting non-homework questions, I might have retagged this ;-) – David Z Dec 31 '11 at 22:55

## 1 Answer

Photons are radiation so their equation of state is $$p = \frac{\rho}{3}$$ where $\rho$ is the energy density. So we have $$p = \frac{mc^2}{3V} = \frac{1\times 9\times 10^{16}\,\,{\rm J}}{0.003\,{\rm m}^3} = 3 \times 10^{19}\,\,{\rm Pa}$$ It's a huge pressure. Not a surprising fact because the actual mass of photons we can produce is negligible. One kilogram worth of photons could be obtained by detonating a reasonable number of H-bombs: the mass would be taken from the difference between the helium and hydrogen nuclear masses.

Note that in this parameterization, the result doesn't depend on the frequency/wavelength/temperature of the photons.

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