Question

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?

Short answer

$\left(a\right)$The total translation kinetic energy of gas molecules,$E_{\mathrm{th}}=1.62\mathrm{kJ}.$

$\left(b\right)$The thermal energy of gas, ${\mathrm{E}}_{\mathrm{th}}=2.7\mathrm{kJ}.$

$\left(c\right)$The speed of molecules,${\mathrm{v}}_{\mathrm{rms}}=1550\mathrm{m}/\mathrm{s}\text{.}$

Step-by-step solution

Step: 1   Finding number of molecules: (part a(

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

Step: 2  Finding the value of Eth: (par b)

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

Step: 3 Finding the mass: (pat c)

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

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

Step: 4 Finding number of mole:

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 $E_{th}$

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

Step: 5 c Finding the root mean square velocity:

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

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