If the kinetic energy of a monoatomic gas mol | vedclass

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(C) The total internal energy $E$ of an ideal gas is given by $E = \frac{f}{2}RT = \frac{f}{2}PV$,where $f$ is the degrees of freedom.For a monoatomic

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$\frac{5}{2}PV$

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(C) The total internal energy $E$ of an ideal gas is given by $E = \frac{f}{2}RT = \frac{f}{2}PV$,where $f$ is the degrees of freedom.For a monoatomic gas,the degrees of freedom $f = 3$. Therefore,$E = \frac{3}{2}PV$.For a diatomic gas at moderate temperatures,the degrees of freedom $f = 5$ ($3$ translational + $2$ rotational).Therefore,the kinetic energy of a diatomic gas molecule is $E = \frac{5}{2}PV$.

If the kinetic energy of a monoatomic gas molecule is $\frac{3}{2}PV$,then the kinetic energy of a diatomic gas molecule is:

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