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A foolball at 25^(@)C has 0.5 mole air m...

A foolball at `25^(@)C` has 0.5 mole air molecules . calculate the internal energy of air in the ball.

Text Solution

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The internal energy of ideal gas `=3/2` NkT. The number of air molecules is given in terms of number of moles so, rewrite the expression as follows:
`U=3/2muRT`
Since `Nk=muR. "Here"mu` is number of moles.
Gas constant `R=8.31 J/("mol")K`
Temperature`T=273+27^(@)C=300K`
`U3/2xx0.5xx8.31xx300=1869.75J`
This is approximately equivalent to the kinetic energy of a man of 57 kg runinf with a speed of 8 `ms^(-1)`
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