2.3 Heat Capacity and Equipartition of Energy

Suppose 2 moles of helium gas at 200 K are mixed with 2 moles of krypton gas at 400 K in a calorimeter. What is the final temperature?

One might think that the internal energy of diatomic gases is given by $E_{\text{int}} = 5 R T / 2 .$ For diatomic gases near room temperature, what additional considerations would cause the internal energy to be less than that? What would cause the internal energy to be more than that? Hint: Their internal energy includes the total energy added in raising the temperature from the boiling point (very low) to room temperature.

To give a helium atom nonzero angular momentum requires about 21.2 eV of energy (that is, 21.2 eV is the difference between the energies of the lowest-energy or ground state and the lowest-energy state with angular momentum). The electron-volt or eV is defined as $1.60 \times 10^{−19} \text{J} .$ Find the temperature T where this amount of energy equals $k_{\text{B}} T / 2 .$ Does this explain why we can ignore the rotational energy of helium for mo…

A sealed, rigid container of 0.560 mol of an unknown ideal gas at a temperature of $30.0 °\text{C}$ is cooled to $−40.0 °\text{C}$. In the process, 980 J of heat are removed from the gas. Is the gas monatomic, diatomic, or polyatomic?

A steel container of mass 135 g contains 24.0 g of ammonia, $\text{NH}_{3}$, which has a molar mass of 17.0 g/mol. The container and gas are in equilibrium at $12.0 °\text{C}$. How much heat has to be removed to reach a temperature of $−20.0 °\text{C}$ ? Ignore the change in volume of the steel and treat the ammonia as an ideal gas with 6 degrees of freedom (3 translational and 3 rotational; at room temperature, the vibrational degree of fre…

Heliox, a mixture of helium and oxygen, is sometimes given to hospital patients who have trouble breathing, because the low mass of helium makes it easier to breathe than air. Suppose helium at $25 °\text{C}$ is mixed with oxygen at $35 °\text{C}$ to make a mixture that is $70 \%$ helium by mole. What is the final temperature? Ignore any heat flow to or from the surroundings, and assume the final volume is the sum of the initial…

In car racing, one advantage of mixing liquid nitrous oxide $\left(\text{N}_{2} \text{O}\right)$ with air is that the boiling of the “nitrous” absorbs latent heat of vaporization and thus cools the air and ultimately the fuel-air mixture, allowing more fuel-air mixture to go into each cylinder. As a very rough look at this process, suppose 1.0 mol of nitrous oxide gas at its boiling point, $−88 °\text{C}$, is mixed with 4.0 mol of air (assumed d…