1.

The average kinetic energy of a gas molecule at 0^(@)C is 5.621xx10^(-27)J. Calculate Boltzman constant. Also calculate the number of molecules present in one mole of the gas.

Answer»

Solution :Average kinetic ENERGY `KE=3/2kT`
`:.k=2/3xx(KE)/^=2/3xx(5.621xx10^(-21)J"MOLECULE"^(-1))/(273K)=1.373xx10^(-23)JK^(-1)"molecule"^(-1)`
No. of molecules in 1 mole of the gas (Avagadro.s no.) `=R/k`
`:.R/k=(8.314JK^(-1)"MOL"^(-1))/(1.373xx10^(-23)JK^(-1)"molecule"^(-1))=6.05xx10^(23)"molecules mol"^(-1)`


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