How many degrees of freedom are associated with `2` gram of helium at NTP ? Calculate the amount of heat energy required to raise the temp. Of this amount from `27^(@) C to 127^(@) C`. Given Boltzmann constant `k_(B) = 1.38 xx 10^(-16)` erg `"molecule"^(-1) K^(-1)` and Avogadro's number `=6.02 xx 10^(23)`.

Explanation

Here, mass of `He=2` gram and its atomic wt. `= 4`
As, No atoms in `4 "gram" = 6-02 xx 10^(23`, No of atoms in `2 "gram"= (6.02 xx 10^(23) xx2)/(4) = 3.01 xx 10^(23)`
As , He is monotomic, each molecule has three degree of freedom.
`:.` Total no. of degree of freedom = ` 3xx 3.01 xx 10^(23)= 9.03 xx 10^(23)`
As energy associated with one degree of freedom per molecule ` = 1/2 K_(B)T`
`:. at T_(1)=(27^(@)C) = 300 K`, heat content of gas
`U_(1)=9-03 xx 10^(23)xx1/2 K_(B)T_(1)=9.03 xx10^(23) xx 1/2 xx 1.38 xx 10^(-23) xx 300J = 1869.2J`
Similarly at `T_(2) = 127^(C) = (127+273)K = 400K`
`U_(2)=9.03 xx 10^(23) xx 1/2 xx 1.38 xx 10^(-23) xx 400J=2492.3 J`
Hence energy required to raise the temp. of gas from `300K "to" 400K` is
`U=U_(2)-U_(1) = 2492.3 - 1869.2 = 623.1 "joule"`.