The rms speed of the molecules in $1.0g$of hydrogen gas is$1800\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ .
a. What is the total translational kinetic energy of the gas molecules?
b. What is the thermal energy of the gas?
c. $500J$of work are done to compress the gas while, in the same process, $1200J$of heat energy are transferred from the gas to the environment. Afterward, what in the rms speed of the molecules?

In the microsystem, the potential energy between the bonds is zero, and a monatomic gas's kinetic energy is translational. As a result, a monatomic gas' thermal energy is $N$atoms is given by equation in the form

$E_{\mathrm{th}}=N{ϵ}_{\mathrm{avg}}$

Where ${ϵ}_{\mathrm{avg}}$ the average amount of energy The mass of the molecule$m$and velocity $v$has a translational kinetic energy on average The average translational kinetic energy of a molecule is affected by its temperature, hence it is connected to the temperature. $T$per molecule in the form

${ϵ}_{\mathrm{avg}}=\frac{1}{2}m{v}_{\mathrm{rms}}^2$

where $k_B$is the Boltzmann's constant and in SI unit its value is

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

Knowing the mass M and the molar mass $m$,we can get the number of molecules by

localid="1648463197023" $N=\frac{M}{m}$.

Using the expressions of $N$and ${ϵ}_{\mathrm{avg}}$in expressions for localid="1648451240549" $v_{\mathrm{rms}}^2$

$\begin{array}{r}{E}_{\mathrm{th}}=N{ϵ}_{\mathrm{avg}}\\ =\frac{M}{m}\frac{1}{2}m{v}_{\mathrm{avg}}^2\\ =\frac{1}{2}M{v}_{\mathrm{rms}}^2\end{array}$

Now,we plug the value for $M$and $v_{\mathrm{rms}}$into equation to get $E_{th}$

localid="1648451415752" $\begin{array}{r}{E}_{\mathrm{th}}=\frac{1}{2}M{v}_{\mathrm{rms}}^2\\ =\frac{1}{2}\left(1\times {10}^{-3}\mathrm{kg}\right)(1800\mathrm{m}/\mathrm{s}{)}^2\\ =1620\mathrm{J}\\ =1.62\mathrm{kJ}.\end{array}$

The potential energy between the bonds is zero in the microsystem and the kinetic energy of a system is translational kinetic energy. So, the thermal energy of a diatomic gas of $n$moles is given by equation in the form

$E_{\mathrm{th}}=\frac{5}{2}nRT$

Where, $n$is the number of moles and $R$is the universal gas constant.

Temperature in terms of root mean square velocity $v_{\mathrm{rms}}$by

$\begin{array}{c}\frac{1}{2}m{v}_{\mathrm{rms}}^2=\frac{3}{2}{k}_{\mathrm{B}}T\\ v_{\mathrm{rms}}^2=\frac{3k_{\mathrm{B}}T}{m}\\ T=\frac{m{v}_{\mathrm{rms}}^2}{3k_{\mathrm{B}}}\end{array}$

The molecular mass of hydrogen is $m=1\mathrm{U}$. But the hydrogen is a diatomic gas $H_2$, so we get the mass for two atoms $m=2u$. Converting this to $kg$we get the mass of one atom of argon by

$m=2\mathrm{u}\times \frac{1.66\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}}=3.32\times {10}^{-27}\mathrm{kg}.$

The values $m,v_{\mathrm{rms}}\text{and}{k}_{\mathrm{B}}$into equation to get $T$

$\begin{array}{r}T=\frac{m{v}_{\mathrm{rms}}^2}{3k_{\mathrm{B}}}\\ =\frac{\left(3.32\times {10}^{-27}\mathrm{kg}\right)(1800\mathrm{m}/\mathrm{s}{)}^2}{3\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)}\\ =260\mathrm{K}.\end{array}$

Knowing the mass $M$and the molar mass,we get the number of moles of the gas by

$n=\frac{M}{m}=\frac{1\mathrm{g}}{2\mathrm{g}/\mathrm{mol}}=0.5\mathrm{mol}$

Now,we plug the values for $n,RandT$into the equation to get $E_{th}$

$\begin{array}{r}{E}_{\mathrm{th}}=\frac{5}{2}nRT\\ =\frac{5}{2}\left(0.5\mathrm{mol}\right)(8.314\mathrm{J}/\mathrm{mol}\cdot \mathrm{K})\left(260\mathrm{K}\right)\\ =2702\mathrm{J}\\ =2.7\mathrm{kJ}.\end{array}$

When the wwork is done on the gas,its energy decreases by $500J$and decreases by $1200J$.so, the final energy of the gas is

$E_{\text{th}}=2702\mathrm{J}+500\mathrm{J}-1200\mathrm{J}=2002\mathrm{J}$

The final temperature by

$\begin{array}{r}{E}_{\mathrm{th}}=\frac{5}{2}nRT\\ T=\frac{2E_{\mathrm{th}}}{5nR}\\ =\frac{2\left(2000\mathrm{J}\right)}{5\left(0.5\mathrm{mol}\right)(8.314\mathrm{J}/\mathrm{mol}\cdot \mathrm{K})}\\ =192\mathrm{K}\end{array}$

Now,we get the root mean square velocity by

$\begin{array}{r}{v}_{\mathrm{rms}}=\sqrt{\frac{3k_{\mathrm{B}}T}{m}}\\ =\sqrt{\frac{3\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(192\mathrm{K}\right)}{\left(3.32\times {10}^{-27}\mathrm{kg}\right)}}\\ =1550\mathrm{m}/\mathrm{s}.\end{array}$