2.3 Heat capacity and equipartition of energy  (Page 4/6)

Experimentally it appears that many polyatomic molecules’ vibrational degrees of freedom can contribute to some extent to their energy at room temperature. Would you expect that fact to increase or decrease their heat capacity from the value R ? Explain.

One might think that the internal energy of diatomic gases is given by $E_{\text{int}}=5RT\text{/}2.$ Do diatomic gases near room temperature have more or less internal energy than that? Hint: Their internal energy includes the total energy added in raising the temperature from the boiling point (very low) to room temperature.

You mix 5 moles of ${\text{H}}_2$ at 300 K with 5 moles of He at 360 K in a perfectly insulated calorimeter. Is the final temperature higher or lower than 330 K?

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 $1.60\times {10}^{\mathrm{-19}}\text{J}\text{.}$ Find the temperature T where this amount of energy equals $k_{\text{B}}T\text{/}2.$ Does this explain why we can ignore the rotational energy of helium for most purposes? (The results for other monatomic gases, and for diatomic gases rotating around the axis connecting the two atoms, have comparable orders of magnitude.)

(a) How much heat must be added to raise the temperature of 1.5 mol of air from $25.0\text{°C}$ to $33.0\text{°C}$ at constant volume? Assume air is completely diatomic. (b) Repeat the problem for the same number of moles of xenon, Xe.

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

A sample of neon gas (Ne, molar mass $M=20.2\text{g/mol})$ at a temperature of $13.0\text{°C}$ is put into a steel container of mass 47.2 g that’s at a temperature of $\mathrm{-40.0}\text{°C}$ . The final temperature is $\mathrm{-28.0}\text{°C}$ . (No heat is exchanged with the surroundings, and you can neglect any change in the volume of the container.) What is the mass of the sample of neon?

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 $\mathrm{-20.0}\text{°C}$ ? Ignore the change in volume of the steel.

A sealed room has a volume of $24{\text{m}}^3$ . It’s filled with air, which may be assumed to be diatomic, at a temperature of $24\text{°C}$ and a pressure of $9.83\times {10}^4\text{Pa}\text{.}$ A 1.00-kg block of ice at its melting point is placed in the room. Assume the walls of the room transfer no heat. What is the equilibrium temperature?

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 volumes.

Professional divers sometimes use heliox, consisting of $79\%$ helium and $21\%$ oxygen by mole. Suppose a perfectly rigid scuba tank with a volume of 11 L contains heliox at an absolute pressure of $2.1\times {10}^7\text{Pa}$ at a temperature of $31\text{°C}$ . (a) How many moles of helium and how many moles of oxygen are in the tank? (b) The diver goes down to a point where the sea temperature is $27\text{°C}$ while using a negligible amount of the mixture. As the gas in the tank reaches this new temperature, how much heat is removed from it?

In car racing, one advantage of mixing liquid nitrous oxide $({\text{N}}_2\text{O})$ 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, $\mathrm{-88}\text{°C}$ , is mixed with 4.0 mol of air (assumed diatomic) at $30\text{°C}$ . What is the final temperature of the mixture? Use the measured heat capacity of ${\text{N}}_2\text{O}$ at $25\text{°C}$ , which is $30.4\text{J/mol}\text{°C}$ . (The primary advantage of nitrous oxide is that it consists of 1/3 oxygen, which is more than air contains, so it supplies more oxygen to burn the fuel. Another advantage is that its decomposition into nitrogen and oxygen releases energy in the cylinder.)