Question

At STP, what is the total translational kinetic energy of the molecules in $1.0\mathrm{mol}$of (a) hydrogen, (b) helium, and (c) oxygen?

Short answer

(A) The energy of Hydrogen $E_{H2}=3403J$

(B) The energy of Helium $E_{He}=3403J$

(C)The energy of Oxygen$E_{O2}=3403J_{}$

Step-by-step solution

Step :1  Molecule with mass and velocity (part a)

(a) The average translational kinetic energy of a molecule with mass$m$ and velocity$v$ is The average translational kinetic energy of a molecule is affected by its temperature, hence it is related to the temperature per molecule in the form.

${ϵ}_{\text{avg}}=\frac{3}{2}{k}_{\mathrm{B}}T$

Where $k_{\mathrm{B}}$is Boltzmann's constant and in $SI$unit its value is

$k_{\mathrm{B}}=1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$

This the energy per molecule. For a number of moles $n$, it contains a number of molecules

$N=n{N}_A$

So, for N number of molecules, the energy in equation (1) will be

${ϵ}_{\mathrm{N}}=N\left(\frac{3}{2}{k}_{\mathrm{B}}T\right)=\frac{3}{2}n{N}_A{k}_{\mathrm{B}}T$

Where $n$is the number of moles and equals $1$mole and $N_A$is Avogadro's number. At STP, the temperature is $T=273\mathrm{K}$. Now, we plug the values for $n,N_A,k_{\mathrm{B}}$and $T$into equation (2) to get the energy for hydrogen

${ϵ}_{{\mathrm{H}}_2}=\frac{3}{2}n{N}_A{k}_{\mathrm{B}}T$

$=\left(\frac{3}{2}\right)\left(1\mathrm{mol}\right)\left(6.023\times {10}^{23}\right)\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(273\mathrm{K}\right)$

$=3403\mathrm{J}$

Step : 2  Energy for helium (part b)

(b) Now, we plug the values for $n,N_A,k_{\mathrm{B}}$and $T$into equation $\left(2\right)$to get the energy for helium

${ϵ}_{{\mathrm{O}}_2}=\frac{3}{2}n{N}_A{k}_{\mathrm{B}}T$

$=\left(\frac{3}{2}\right)\left(1\mathrm{mol}\right)\left(6.023\times {10}^{23}\right)\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(273\mathrm{K}\right)$

$=3403\mathrm{J}$

Step :3 Energy for oxygen (part c)

(c) Plug the values for $n,N_A,k_B$and $T$into equation (2) to get the energy for oxygen

${ϵ}_{{\mathrm{O}}_2}=\frac{3}{2}n{N}_A{k}_{\mathrm{B}}T$

$=\left(\frac{3}{2}\right)\left(1\mathrm{mol}\right)\left(6.023\times {10}^{23}\right)\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(273\mathrm{K}\right)$

$=3403\mathrm{J}$

As shown, the transitional energy depends on the temperature only not the mass of the gas. So, the three gasses have the same transitional kinetic energy for the same number of moles.