PV=1/3 E connecting the “light” radiation energy E in a box of volume V to the r

Question

PV=1/3 E connecting the “light” radiation energy E in a box of volume V to the radiation pressure P inside the box. E is the internal energy of the radiation which we will call U, i.e. E = U. Defining the radiation energy density u =U/V, we have and from PV=1/3E , we get P=1/3u . These equations in boxes provide a key to understanding the thermodynamics of radiation energy.

a)If we hold the temperature constant what effect does doubling the volume of the container have on the pressure? The internal energy U? If we double the temperature what effect does this have on the pressure? Internal energy?

(b) A box 0.1 m3 contains 1 mole of a monatomic ideal gas. The entire box and contents are held at 300K. Determine the ratio U(radiation)/U(gas) of the total radiation energy in the box U(radiaion) to the energy of the ideal gas Ugas.
(c) At 300K determine the ratio of P(radiation)/ P(gas)?
(d) At what temperature is U(radiation)/ U(gas) = 1 ? At what temperature is P(radiation)/ P(gas) =1? Where might you find these temperatures?

Explanation / Answer

a)if temperature is constant E=U is constant.so,doubling V halves P.Doubling temperature doubles U but V remains constant.So,P doubles.

c)=1/3 becuse P(radiation)V=1/3nRT(becuse E=nRT),P(gas)V=nRT

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