Problem 39 - Chemistry: The Central Science

Question

A scuba diver's tank contains \(0.29 \mathrm{~kg}\) of \(\mathrm{O}_{2}\) compressed into a volume of \(2.3 \mathrm{~L}\) (a) Calculate the gas pressure inside the tank at \(9^{\circ} \mathrm{C}\). (b) What volume would this oxygen occupy at \(26^{\circ} \mathrm{C}\) and \(0.95 \mathrm{~atm}\) ?

Step-by-Step Solution

Answer

The gas pressure inside the scuba diver's tank at \(9^{\circ} \mathrm{C}\) is approximately 97.4 atm. The volume of the oxygen at \(26^{\circ} \mathrm{C}\) and \(0.95 \mathrm{~atm}\) is approximately 233.6 L.

Step 1: Find moles of oxygen(O2)

To find the amount of oxygen in moles, use the molar mass of O2 which is approximately 32 g/mol: \[n=\frac{mass}{molar\: mass}=\frac{0.29 \mathrm{~kg}}{32 \mathrm{~g/mol}}\] Convert kg to g: \[\frac{0.29 \mathrm{~kg}}{1}*\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}}=290 \mathrm{~g}\] \[n=\frac{290 \mathrm{~g}}{32 \mathrm{~g/mol}}\approx 9.06\, \text{moles}\]

Step 2: Convert temperatures to Kelvin

Both temperatures need to be converted to Kelvin, as the ideal gas law and combined gas formulas require them in that unit. Initial temperature: \(T_1 = 9^{\circ} \mathrm{C} + 273.15 \approx 282.15 \, \text{K}\) Final temperature: \(T_2 = 26^{\circ} \mathrm{C} + 273.15 \approx 299.15 \, \text{K}\)

Step 3: Calculate the gas pressure inside the tank (part a)

Use the ideal gas law formula to find the pressure: \[PV=nRT\] \[P=\frac{nRT}{V}=\frac{9.06 \, \text{moles} * 0.0821 \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}} * 282.15 \, \text{K}}{2.3 \, \text{L}} \approx 97.4 \, \text{atm}\] The gas pressure inside the scuba diver's tank is approximately 97.4 atm.

Step 4: Calculate the volume of the oxygen at the new conditions (part b)

Use the combined gas law formula and the given data to find the new volume: \[\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}\] Plug in the given and calculated values: \[\frac{97.4 \, \text{atm} * 2.3 \, \text{L}}{282.15 \, \text{K}} = \frac{0.95 \, \text{atm} * V_{2}}{299.15 \, \text{K}}\] Now solve for \(V_2\): \[V_{2} = \frac{97.4 \, \text{atm} * 2.3 \, \text{L} * 299.15 \, \text{K}}{0.95 \, \text{atm} * 282.15 \, \text{K}} \approx 233.6 \, \text{L}\] The volume of the oxygen at \(26^{\circ} \mathrm{C}\) and \(0.95 \mathrm{~atm}\) is approximately 233.6 L.

Key Concepts

Moles of Gas

Understanding the concept of moles is essential when dealing with gases. The mole is a unit used to measure the amount of substance. One mole of any substance, like the oxygen in our scuba tank example, contains Avogadro's number of molecules, which is approximately \(6.022 \times 10^{23}\) molecules. This quantity allows chemists to link the microscopic scale of atoms and molecules to the macroscopic scale we measure in the lab. To find the moles from a given mass, you divide the mass of the substance by its molar mass. For example, our scuba diver's tank contains 0.29 kg of oxygen. By converting this to grams (290 g) and dividing by the molar mass of oxygen (32 g/mol), we obtain:

• 290 g / 32 g/mol = 9.06 moles of \(O_2\).

This conversion from mass to moles is a fundamental skill in chemistry, especially when using the ideal gas law.

Gas Pressure Calculation

Gas pressure is the force that the gas exerts on the walls of its container. It's calculated using the ideal gas law, which connects pressure (\(P\)), volume (\(V\)), and temperature (\(T\)) through the formula: \[ PV = nRT \] where \(R\) is the universal gas constant, and \(n\) is the number of moles. By rearranging this formula to solve for pressure, we have: \[ P = \frac{nRT}{V} \] In our scuba diver problem, substituting the known values: 9.06 moles of \(O_2\), \(R = 0.0821 \frac{L \, atm}{mol \, K}\), \(T = 282.15 \, K\), and \(V = 2.3 \, L\), gives us the pressure inside the tank. It calculates to approximately 97.4 atm.

• This outcome reveals how significantly pressure rises inside a compressed gas tank.
• It's also a reminder of why understanding gas behavior is crucial for safety in situations like scuba diving.

Temperature Conversion

When using gas equations, temperature should always be in Kelvin. The Kelvin scale is an absolute measure of temperature, meaning it starts at absolute zero, the point where molecular motion stops. To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature.

• For example, \(9^{\circ} C\) becomes: \(9 + 273.15 = 282.15 \, K\).
• Similarly, \(26^{\circ} C\) becomes: \(26 + 273.15 = 299.15 \, K\).

Using Kelvin helps ensure accuracy when applying the ideal or combined gas laws. It is crucial because these laws are based on the behavior of particles at absolute temperatures, which must be accounted for accurately.

Combined Gas Law

The combined gas law combines three individual gas laws into one equation: Boyle's Law, Charles's Law, and Gay-Lussac's Law. It takes the form: \[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \] This law is particularly useful when dealing with changes in conditions like pressure, volume, and temperature for a fixed amount of gas. In our scuba diver example, we use the combined gas law to find out the volume of oxygen at a new pressure and temperature: 0.95 atm and 299.15 K. Using the relation between initial and final states:

• \(P_1 = 97.4, V_1 = 2.3 \, L, T_1 = 282.15 \, K\)
• \(P_2 = 0.95, T_2 = 299.15 \, K\)

We solve for \(V_2\), which represents the new volume under these conditions. This equation is vital in practical applications, helping predict how gases will behave when subjected to different environments.