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12.Kinetic Theory of Gases
normal
The translational kinetic energy of gas molecule for one mole of the gas is equal to
A
$\frac{3}{2}RT$
B
$\frac{2}{3}RT$
C
$\frac{1}{2}RT$
D
$\frac{2}{3}KT$
Solution
The transation kinetic energy pee molecule:
$\frac{K L}{m}=\frac{3}{2} . K_BT$
were, $m =$ no. of molecules , $n=$ moles, $N=$ avagardo's number
also, $n=\frac{m}{N}$
so, $\frac{K F}{n N}=\frac{3}{2} K_ B T$
$\frac{K E}{n}=\frac{3}{2} N K_B T$
as $N K_B=R$ (gas constant)
KE molar $=\frac{3}{2} R T$
Standard 11
Physics
