Problem 41 - 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 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 atm?

Step-by-Step Solution

Answer

(a) The gas pressure inside the scuba diver's tank at \(9^\circ C\) is 10.15 atm. (b) At \(26^\circ C\) and 0.95 atm, the oxygen would occupy a volume of 29.4 L.

Step 1: Calculate the number of moles of Oxygen

First, we need to find the number of moles of Oxygen in the tank. To do this, we'll use the mass of Oxygen and its molar mass. The molar mass of O2 is 32 g/mol. Number of moles (n) = \(\frac{Mass}{Molar\ mass}\) n = \(\frac{0.29 kg}{32 g/mol}\) (1 kg = 1000 g) n = \(\frac{290 g}{32 g/mol}\) = \(9.06 mol\)

Step 2: Convert temperature to Kelvin

We have to work with temperature in Kelvin while dealing with the gas law formula. To convert a Celsius temperature to Kelvin, we add 273.15 to the Celsius temperature. T1 (initial temperature) = \(9^\circ C + 273.15\) = \(282.15 K\) T2 (final temperature) = \(26^\circ C + 273.15\) = \(299.15 K\)

Step 3: Calculate pressure inside the tank

Now, we'll use the ideal gas law formula to find the pressure inside the tank: \(PV = nRT\) We know the number of moles (n) and the initial temperature (T1). We also know the volume is 2.3 L, but we need to convert it to cubic meters (SI unit): V = 2.3 L × \(\frac{1 m^3}{1000 L}\) = \(2.3 × 10^{-3} m^3\) Now we can find the pressure: P = \(\frac{nRT}{V}\) Using R = \(8.314 J/(mol·K)\), we get: P = \(\frac{9.06 mol × 8.314 J/(mol·K) × 282.15 K}{2.3 × 10^{-3} m^3}\) = \(1028651.87 Pa\) We can convert the pressure to atm by using the conversion factor \(1 atm = 101325 Pa\): P = \(\frac{1028651.87 Pa}{101325 Pa/atm}\) = 10.15 atm

Step 4: Calculate the final volume

We need to find the final volume of the oxygen gas at T2 and 0.95 atm. For this, we can use the combined gas law formula, which is derived from the ideal gas law and expresses the relationship between the initial and final states of a gas: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) We know \(P_1\), \(V_1\), \(T_1\), \(P_2\), and \(T_2\), so we can solve for \(V_2\): \(V_2 = \frac{P_1V_1T_2}{P_2T_1}\) \(V_2 = \frac{10.15 atm × 2.3 × 10^{-3} m^3 × 299.15 K}{0.95 atm × 282.15 K}\) \(V_2 = 0.0294 m^3\) Now, convert the volume back to liters: \(V_2 = 0.0294 m^3 × \frac{1000 L}{1 m^3}\) = 29.4 L The final volume of the oxygen at \(26^\circ C\) and 0.95 atm is 29.4 L.

Key Concepts

Molar Mass

Understanding the concept of molar mass is crucial when dealing with gases and their reactions. Molar mass, often expressed in units of grams per mole (g/mol), is the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. For diatomic oxygen, \( ext{O}_2\), the molar mass is calculated by multiplying the atomic mass of an oxygen atom (approximately 16 g/mol) by 2. Thus, the molar mass of \( ext{O}_2\) is 32 g/mol.
In this exercise, we have a mass of \(0.29 ext{ kg} = 290 ext{ g}\) of \( ext{O}_2\). To find out how many moles of oxygen are in the scuba diver's tank, we use the formula:

• Number of moles \(n = \frac{290 ext{ g}}{32 ext{ g/mol}}\)

By performing this calculation, we find there are approximately 9.06 moles of oxygen in the tank.
Knowing the number of moles enables us to explore further properties of the gas, such as pressure and volume under different conditions.

Combined Gas Law

The Combined Gas Law is an essential tool for understanding how changes in pressure, volume, and temperature affect a sample of gas. It combines three fundamental gas laws - Boyle's Law, Charles's Law, and Gay-Lussac's Law - into a single relationship:

• \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\)

Here, \(P\), \(V\), and \(T\) represent the pressure, volume, and temperature of the gas, respectively. The subscript '1' refers to the initial state, while '2' refers to the final state. This equation assumes the amount of gas remains constant.
In our exercise, the Combined Gas Law helps us find the new volume of oxygen at a different temperature (from 9°C to 26°C) and pressure (from 10.15 atm to 0.95 atm). By substituting the known values into the equation, we solve for the final volume \(V_2\). This demonstrates how the properties of gases change across different conditions.
Remember, temperature must always be in Kelvin for these calculations. To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature.

Gas Pressure Calculation

Gas pressure calculation heavily relies on the ideal gas law, which is a fundamental equation that describes the behavior of gases. The equation is given by:

• \(PV = nRT\)

where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.
For the scuba diver's tank example, we need to calculate the pressure inside the tank. With 9.06 moles of oxygen, a tank volume of 2.3 L (converted to cubic meters for SI units), and a temperature of 282.15 K, we can rearrange the ideal gas law to solve for pressure:

• \(P = \frac{nRT}{V}\)

Using the constant \(R = 8.314 ext{ J/(mol·K)}\), the calculation provides a pressure of approximately 10.15 atm after converting from Pascals. This highlights the importance of unit conversion, as pressure is commonly expressed in atmospheres in practical applications.
Understanding these calculations allows for better handling of gases under various conditions, ensuring safety and efficiency in applications such as scuba diving.