Question

Calculate the ratio of total internal energy of H and He at 300k?

Answer 1

$\huge\underline{\underline{\bf \blue{Question-}}}$

Calculate the ratio of total internal energy of H and He at 300k?

$\huge\underline{\underline{\bf \blue{Solution-}}}$

✪$\large\underline{\underline{\mathfrak \red{Note-}}}$

Helium (He) is Monoatomic gas and for Monoatomic gas degree of freedom at all temperature is 3

Hydrogen ${\sf H_2}$ is diatomic gas , and for diatomic gas degree of freedom at low temperature is 5 and at high temperature is 7

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300 K is low temperature ,

Therefore ,

$\large{\boxed{\bf \red{U=\dfrac{degree\:of\: freedom}{2}kT} }}$

$\implies{\sf U_{He}=\dfrac{3}{2}kT }$

$\implies{\sf U_{H_2}=\dfrac{5}{2}kT }$

❏ Ratio ➝

$\implies{\sf \dfrac{U_{He}}{U_{H_2}}=\dfrac{3/2k×300}{5/2k×300}}$

$\implies{\sf \dfrac{U_{He}}{U_{H_2}}=\dfrac{3}{5} }$

$\implies{\bf \red{U_{He}:U_{H_2}=3:5} }$

$\huge\underline{\underline{\bf \blue{Answer-}}}$

Ratio of internal energy is ${\bf \red{U_{He}:U_{H_2}=3:5} }$

Answer 2

The ratio of their ratio of total internal energy of H and He at 300  K is 5 : 3 .

We need to find the ratio of total internal energy of H and He at 300 K .

We know , Internal energy is given by :

$U=\dfrac{nkT}{2}$ { Here n is degree of freedom }

Now , $H_2$ is diatomic therefore ( In given question H is given but hydrogen is always present in  $H_2$ state ) , its degree of freedom is 5 and He is mono atomic therefore , its degree of freedom is 3 .

Therefore , their ratio is :

$\dfrac{H_2}{He}=\dfrac{\dfrac{5kT}{2}}{\dfrac{3kT}{2}}\\\\\dfrac{H_2}{He}=\dfrac{5}{3}$

Therefore , the ratio of their ratio of total internal energy of H and He at 300  K is 5 : 3 .

Hence , this is the required solution .

Learn More :

Thermodynamics

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