Example 12.2-1

The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft³), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon?

Given: volume, temperature, and pressure

Asked for: amount of gas

Strategy:

A Solve the ideal gas law for the unknown quantity, in this case n.

B Make sure that all quantities are given in units that are compatible with the units of the gas constant. If necessary, convert them to the appropriate units, insert them into the equation you have derived, and then calculate the number of moles of hydrogen gas needed.

Solution:

A We are given values for P, T, and V and asked to calculate n. If we solve the ideal gas law (Equation 12.2(eq4)) for n, we obtain

$$n=\frac{PV}{RT}$$

B P and T are given in units that are not compatible with the units of the gas constant [R = 0.082057 (L·atm)/(K·mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres:

$$\begin{array}{c}P=(745\cancel{\text{mmHg}})\left(\frac{1\text{atm}}{760\cancel{\text{mmHg}}}\right)=0.980\text{atm}\\ V=\mathrm{31,150}\text{L}\left(\text{given}\right)\\ T=30+273=303\mathrm{K}\end{array}$$

Substituting these values into the expression we derived for n, we obtain

$$n=\frac{PV}{RT}=\frac{(0.980\cancel{\text{atm}})(\mathrm{31,150}\cancel{\text{L}})}{[0.082057\left(\cancel{\text{L}}·\cancel{\text{atm}}\right)/\left(\cancel{\text{K}}·\text{mol}\right)](303\cancel{\text{K}})}=1.23\times {10}^3{\text{mol}}_{}$$

Exercise

Suppose that an "empty" aerosol spray-paint can has a volume of 0.406 L and contains 0.025 mol of a propellant gas such as CO₂. What is the pressure of the gas at 25°C?

Answer: 1.5 atm

Example 12.2-2

Calculate the density of butane at 25°C and a pressure of 750 mmHg.

Given: compound, temperature, and pressure

Asked for: density

Strategy:

A Calculate the molar mass of butane and convert all quantities to appropriate units for the value of the gas constant.

B Substitute these values into Equation 12.2(eq9) to obtain the density.

Solution:

A The molar mass of butane (C₄H₁₀) is

(4)(12.011)+(10)(1.0079)=58.123 g/mol

Using 0.082057 (L·atm)/(K·mol) for R means that we need to convert the temperature from degrees Celsius to kelvins (T=25+273=298 K) and the pressure from millimeters of mercury to atmospheres:

$$(750\cancel{\text{mmHg}})\left(\frac{1\text{atm}}{760\cancel{\text{mmHg}}}\right)=0.987\text{atm}$$

B Substituting these values into Equation 12.2(eq9) gives

$$d=\frac{PM}{RT}=\frac{(0.987\cancel{\text{atm}})(58.123\text{g/}\cancel{\text{mol}}\text{)}}{\left[0.082057\text{(L}·\cancel{\text{atm}}\text{)/}\left(\cancel{\text{K}}·\cancel{\text{mol}}\right)\right](298\cancel{\text{K}})}=2.35\text{g/L}$$

Exercise

Radon (Rn) is a radioactive gas formed by the decay of naturally occurring uranium in rocks such as granite. It tends to collect in the basements of houses and poses a significant health risk if present in indoor air. Many states now require that houses be tested for radon before they are sold. Calculate the density of radon at 1.00 atm pressure and 20°C and compare it with the density of nitrogen gas, which constitutes 80% of the atmosphere, under the same conditions to see why radon is found in basements rather than in attics.

Answer: radon, 9.23 g/L; N₂, 1.17 g/L

Example 12.2-3

The reaction of a copper penny with nitric acid results in the formation of a red-brown gaseous compound containing nitrogen and oxygen. A sample of the gas at a pressure of 727 mmHg and a temperature of 18°C weighs 0.289 g in a flask with a volume of 157.0 mL. Calculate the molar mass of the gas and suggest a reasonable chemical formula for the compound.

Given: pressure, temperature, mass, and volume

Asked for: molar mass and chemical formula

Strategy:

A Solve Equation 12.2(eq9) for the molar mass of the gas and then calculate the density of the gas from the information given.

B Convert all known quantities to the appropriate units for the gas constant being used. Substitute the known values into your equation and solve for the molar mass.

C Propose a reasonable empirical formula using the atomic masses of nitrogen and oxygen and the calculated molar mass of the gas.

Solution:

A Solving Equation 12.2(eq9) for the molar mass gives

$$M=\frac{dRT}{P}$$

Density is the mass of the gas divided by its volume:

$$d=\frac{m}{V}=\frac{0.289\text{g}}{0.157\text{L}}=1.84\text{g/L}$$

B We must convert the other quantities to the appropriate units before inserting them into the equation:

$$\begin{array}{c}T=18+273=291\text{K}\\ P=(727\cancel{\text{mmHg}})\left(\frac{1\text{atm}}{760\cancel{\text{mmHg}}}\right)=0.957\text{atm}\end{array}$$

The molar mass of the unknown gas is thus

$$M=\frac{dRT}{P}=\frac{(1.84\text{g/}\cancel{\text{L}})[0.082057\left(\cancel{\text{L}}·\cancel{\text{atm}}\right)/(\cancel{\text{K}}·\text{mol)}](291\cancel{\text{K}})}{0.957\cancel{\text{atm}}}=45.9\text{g/mol}$$

C The atomic masses of N and O are approximately 14 and 16, respectively, so we can construct a list showing the masses of possible combinations:

$$\begin{array}{c}\text{NO}=\text{14}+\text{16}=\text{30g/mol}\\ {\text{N}}_2\text{O}=\left(2\right)\left(14\right)+16=44\text{g/mol}\\ {\text{NO}}_2=14+\left(2\right)\left(16\right)=46\text{g/mol}\end{array}$$

The most likely choice is NO₂ which is in agreement with the data. The red-brown color of smog also results from the presence of NO₂ gas.

Exercise

You are in charge of interpreting the data from an unmanned space probe that has just landed on Venus and sent back a report on its atmosphere. The data are as follows: pressure, 90 atm; temperature, 557°C; density, 58 g/L. The major constituent of the atmosphere (>95%) is carbon. Calculate the molar mass of the major gas present and identify it.

Answer: 44 g/mol; CO₂