91Ó°ÊÓ

a) The thermal energy can be represented as since the gases are monoatomic.

$E_{th}=\frac{3}{2}nRT$

The number of moles can be expressed using the ideal gas law.

$pV=nRT⇒n=\frac{pV}{RT}$

The thermal energy will be

$E_{th}=\frac{3}{2}pV$

In our cases, we have

$E_{He}=\frac{3}{2}·\left(2×{10}^5\right)·\left(100×{10}^{-6}\right)=30.4\mathrm{J}$

$E_{Ar}=\frac{3}{2}·\left(4×{10}^5\right)·\left(200×{10}^{-6}\right)=121.6\mathrm{J}$

B) We may calculate the ratio of the number of moles of the two gases using the ideal gas law's equation for the number of moles.

$\frac{n_{He}}{n_{Ar}}=\frac{0.5·0.5}{0.5}=0.5$

That example, there are two times as many argon molecules as there are helium molecules.

The temperature as a function of thermal energy may be written as

$T=\frac{2E_{th}}{3nR}$

Both gases will have the same temperature when they are in thermal equilibrium, which means the ratio of their thermal energy to their mole numbers will be the same. That is

$\frac{E_{He}}{n_{He}}=\frac{E_{Ar}}{n_{Ar}}$

This indicates that the final energy of the argon molecules will be twice that of the helium molecules in our situation. The sum of the two initial energies is the final total energy, according to the law of conservation of energy. As a result, we can locate

$E_{tot}=E_{he}+E_{ar}$

$=30.4+121.6$

$=152J$

This means that the final energy of the argon will be $121.6J$, while the final energy of the helium will be $30.4J$.

Next step is to determine the number of moles of argon gas.

$n_{ar}=\frac{P_{ar}{V}_{ar}}{R{T}_{ar}}$

$=\frac{(4.0)\left(200\right)}{(8.31)\left(673\right)}$

$=0.0145\mathrm{mol}$

After that, we need to determine the number of moles of helium gas.

$n_{he}=\frac{P_{he}{V}_{he}}{R{T}_{he}}$

$=\frac{(2.0)\left(100\right)}{(8.31)\left(373\right)}$

$=0.00654\mathrm{mol}$

c) We can calculate the numerical value of the energy change using simple subtraction. Because argon has a higher temperature than helium, the energy was transferred from the former to the latter.

$\mathrm{Δ}E=E_{h{e}_f}-E_{h{e}_i}$

$=\frac{n_{he}}{n_{he}+n_{ar}}{E}_{tot}-E_{h{e}_i}$

$=\frac{0.00654}{0.00654+0.0145}\left(152\right)-30.4$

$=16.8\mathrm{J}$

d) Knowing that the ratio of the number of moles was $2$to $1$, the temperature change also follows this pattern. This means that the increase in temperature for the helium was twice the decrease for the argon.

$E_{he}=\frac{3}{2}{n}_{he}RT$

We derive the numerical value of the final temperature by substituting values into the preceding equation.

$47.3=\frac{3}{2}(0.00654)(8.31)T$

When we simplify above relation for the final temperature, we get values of the:localid="1648442529072" $T=580\mathrm{K}\mathrm{or}307°\mathrm{C}$

e) By rearranging the formula that connects thermal energy to pressure, we arrive at

$p=\frac{2E_{th}}{3V}$-----------(1)

We can calculate the final pressures using the final energy and volumes.

$p_{He}=\frac{2E_{He}}{3V_{He}}=\frac{2·50}{3·100×{10}^{-6}}=3.11\mathrm{atm}$

$p_{Ar}=\frac{2E_{Ar}}{3V_{Ar}}=\frac{2·100}{3·200×{10}^{-6}}=3.45\mathrm{atm}$