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The radius of a molecule of Argon gas is $ 1.78 A ^\circ C$ . Find the mean free path of molecules of Argon at 0° C temperature and 1 atm pressure. $ k_B = 1.38 \times 10^{-23} JK^{-1} $
$ \bar l = { 1 \over \sqrt 2 \pi n d^2 } = {k_B T \over \sqrt 2 \pi Pd^2 } = 6.65 \times 10^{-8} m $
A monoatomic gas molecule has
A monoatomic gas moecule has only 3 translational degrees of freedom
A diatomic molecule has how many degrees of freedom (For rigid rotator)
A diatomic molecule has 3 translational and 2 rotational degrees of freedom. Hence total degrees of freedom , f = 3 + 2 = 5
The degrees of freedom for triatomic gas is (At room temperature)
For a triatomic gas f = 6 ( 3 translation + 3 rotational )
If the degrees of freedom of a gas are f, then the ratio of two specific heats $ { C_p \over C_v}$ is given by
$ { C_p \over C_v } = \gamma = 1 + { 2 \over f } $
A diatomic gas molecule has translational, rotational and vibrational degrees of freedom. The $ { C_p \over C_v}$ is
Degrees of freedom = 3 ( translatory ) + 2 ( rotatary ) + 1 ( vibratory_ = 6 $ { C_p \over C_v} = \gamma = 1 + { 2 \over f } = 1 + { 2 \over 6 } = 1 + { 1 \over 3} = { 4 \over 3 } = 1.33 $
The value of $C_v$ for one mole of neon gas is
Neon gas is mono atomic and for mono atomic gases $ C_v = {3 \over 2 } R $
The relation between two specific heats of a gas is
When $ C_p $ and $ C _ V$ are given with calorie and R with Joule then $ C_p - C_v = R/J $
The molar specific heat at constant pressure for a monoatomic gas is
$ C_p -C_v = R \Rightarrow C_p =R + C_v \Rightarrow R + { f \over 2} R = R + {3 \over 2 } R = { 5 \over 2} R $ ( f = 3 )
For a gas $ { R \over C_v } = 0.67 $ .This gas is made up of molecules which are
$C_v = { R \over 0.67 } = 1.5 R = {3 \over 2 } R $ This is the value for mono atomic gases
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