For H_2, O_2,and He gases at the same tempera | vedclass

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(A) According to the ideal gas equation,$PV = nRT$,where $n$ is the number of moles.Given that $P, V, R$,and $T$ are the same for all three gases,the

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$H_2$

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(A) According to the ideal gas equation,$PV = nRT$,where $n$ is the number of moles.Given that $P, V, R$,and $T$ are the same for all three gases,the number of moles $n = \frac{PV}{RT}$ is the same for all.Number of molecules $N = n 	imes N_A$,where $N_A$ is Avogadro's number. Thus,the number of molecules is the same for all.Now,we calculate the number of atoms:For $H_2$ (diatomic): Number of atoms $= 2 	imes n 	imes N_A = 2nN_A$.For $O_2$ (diatomic): Number of atoms $= 2 	imes n 	imes N_A = 2nN_A$.For $He$ (monoatomic): Number of atoms $= 1 	imes n 	imes N_A = nN_A$.Comparing these,$H_2$ and $O_2$ have the same number of atoms,which is higher than that of $He$.

For $H_2, O_2$,and $He$ gases at the same temperature and pressure,the volumes are equal. Which of the following has the highest number of atoms?

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