Question

$2.0g$of helium at an initial temperature of $300K$interacts thermally with $8.0g$of oxygen at an initial temperature of $600K$.

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?

Short answer

(a) The initial thermal energy of gas is $1870\mathrm{J},6240\mathrm{J}$

(b)The final thermal energy of gas is $3040\mathrm{J},5070\mathrm{J}$

(c)The heat energy is $1170\mathrm{J}$(d)The final temperature is $488\mathrm{K}$

Step-by-step solution

Introduction (part a)

a. The initial energy can be found as

$E=\frac{i}{2}nRT=\frac{i}{2}\frac{m}{M}RT$

where $i$is the number of degrees of freedom, $m$is the mass of molecules and $M$denotes their molecular weight. Setting the two latter in grams, we don't have to do any unit conversion. Let us also, for simplicity, denote the indexes of helium by $1$and those of oxygen by $2$. Therefore, we will have

$E_1=\frac{3}{2}·\frac{2}{4}·8.314·300=1870\mathrm{J}$

$E_2=\frac{5}{2}·\frac{8}{32}·8.314·600=3118\mathrm{J}$

The total energy of the system (part b)

b. The total energy of the system will be

$E=E_1+E_2=\underset{¯}{8110\mathrm{J}}$

For the sake of simplicity, and because the numbers are quite "beautiful," let's just suppose that$n_1=\frac{m_1}{M_1}=0.5$and $n_2=\frac{m_2}{M_2}$.

Therefore, we will not consider the number of moles to be equal in order to demonstrate how we would proceed if they were not.

The total energy of the system when both sides are in thermal equilibrium, which we have determined using the equation of conservation of energy, will be

$E=\left(i_1{n}_1+i_2{n}_2\right)\frac{1}{2}RT$

This means that the final distribution of energy is

$E_{1f}=\frac{i_1{n}_1}{i_1{n}_1+i_2{n}_2}E=\frac{3}{3+5}·8110=3040\mathrm{J}$

$E_{2f}=\frac{i_2{n}_2}{i_1{n}_1+i_2{n}_2}E=\frac{5}{3+5}·8110=5070\mathrm{J}.$

The heat transformation (part c)

c. The thermal transmission is symmetric, meaning that the amount of heat given by one side is received by the other. As a result, we can calculate it at any of them, even if it's helium:

$Q=\Delta E=3040-1870=1170\mathrm{J}$

The final temperature value  (part d)

d. The final temperature can also be easily found in either of the gases, let we calculate in in the first one. Knowing the energy, the temperature will be

$T=\frac{2E_f}{inR}=\frac{2·3040}{3·0.5·8.314}=488\mathrm{K}$