Calculate the total number of degress of freedom possessed by the molecules in one ` cm^3` of ` H_2 ` gas at NTP .

Explanation

22400 `cm^2` of every gas contains `6.02 xx 10^(23)` molecules
` therefore ` Number of molecules in `1 cm ^3 ` of ` H_2 ` gas `=( 6.02 xx 10^(23))/( 22400 ) = 0.26875 xx 10^(20)`
Number of degrees of freedom of a` H_2` gas molecule = 5
` therefore ` Total number of degress of freedom of `0.26875 xx 10^(20)` molecules
`= 0.26875 xx 10^(20) xx 5 = 1.34 375 xx 10^(20)`