Question

A $100c{m}^3$box contains helium at a pressure of $2.0atm$and a temperature of $100°C$. It is placed in thermal contact with a$200c{m}^3$box containing argon at a pressure of$4.0atm$and a temperature of $400°C$.

a. What is the initial thermal energy of each gas?

b. What is the final thermal energy of each gas?

c. How much heat energy is transferred, and in which direction?

d. What is the final temperature?

e. What is the final pressure in each box?

Short answer

(A)The thermal energy of

$E_{He}=30.4J$

$E_{Ar}=121.6J$

(B)The final energy of helium is $152J$

(C)The higher temperature of helium $16.8J$

(D) The final temperature is$T=580kor307°C$

(E)The final pressure value of

localid="1648442827166" $p_{He}=3.11\mathrm{atm}$

localid="1648442839674" $p_{Ar}=3.45\mathrm{atm}$

Step-by-step solution

Introduction (part a)

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.

$pV=nRT\Rightarrow n=\frac{pV}{RT}$

The thermal energy will be

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

In our cases, we have

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

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

Expression of gas law (part b)

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 $30.4J$.

Substitution 

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.

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

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

$=0.00654\mathrm{mol}$

Expression of helium gas (part c)

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.

$\Delta E=E_{h{e}_f}-E_{h{e}_i}$

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

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

$=16.8\mathrm{J}$

Find the temperature value  (part d)

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" $T=580\mathrm{K}\mathrm{or}307°\mathrm{C}$

Final pressure (part e)

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.

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

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