1.

Calculate the number of degree of freedom in 15 c.c. Of nitrogen at N.T.P.?

Answer» `V=15 c.c. N=?`
At N.T.P., number of nitrogen molecules in `22400 c.c. = 6.02 xx 10^(23)`
`:.` number of nitrogen molecules in 15 c.c.
`=(6.02 xx 10^(23) xx 15)/(22400)`
As number of degrees of freedom of each diatomic mol. = 5
`:.` Total no. of degree of freedom
`=(6.02 xx 10^(23) xx 15 xx 5)/(22400) = 2.015 xx 10^(21)`.


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