Problem 39 - Chemistry: The Central Science in SI Units, Global Edition

Question

A scuba diver's tank contains \(2.50 \mathrm{~kg}\) of \(\mathrm{O}_{2}\) compressed into a volume of \(11.0 \mathrm{~L}\). (a) Calculate the gas pressure inside the tank at \(10^{\circ} \mathrm{C}\). (b) What volume would this oxygen occupy at \(25^{\circ} \mathrm{C}\) and \(101.33 \mathrm{kPa} ?\)

Step-by-Step Solution

Answer

The gas pressure inside the tank is 166.5 kPa, and the oxygen would occupy a volume of 1.90 L at \(25^{\circ}C\) and \(101.33\:kPa\).

Step 1: (Step 1) Converting given values to appropriate units.

Before we use the ideal gas law, we must convert the given values into appropriate units. The mass of oxygen is given in kg, which should be converted to moles (n) using the molar mass of oxygen which is approximately 32 g/mol. The temperature should be converted to Kelvin (K) for the ideal gas law calculation. \(mass\_of\_O2 = 2.50\:kg\) \(volume\_of\_tank = 11.0\:L\) To convert the mass of oxygen to moles : \(moles = \frac{mass}{molar\_mass}\) where \(molar\_mass\_of\_O2 = 32\:g/mol\)

Step 2: (Step 2) Calculate the number of moles of oxygen

First, we need to convert the mass of oxygen from kg to grams (1 kg = 1000 g). \(mass\_of\_O2 = 2.50\:kg * 1000\:g/kg = 2500\:g\) Next, we will calculate the number of moles of oxygen (n) using the molar mass of oxygen. \(moles = \frac{mass\_of\_O2}{molar\_mass\_of\_O2}\) \(moles = \frac{2500\:g}{32\:g/mol} = 78.125 \:mol\)

Step 3: (Step 3) Convert temperature to Kelvin

To use the ideal gas law, we need to convert the temperature from Celsius to Kelvin. \(Tem_1 = 10^{\circ}C\) \(Tem_2 = 25^{\circ}C\) Therefore, \(Tem_1\_K = Tem_1 + 273.15 = 283.15\:K\) \(Tem_2\_K = Tem_2 + 273.15 = 298.15\:K\)

Step 4: (Step 4) Calculate the gas pressure inside the tank

Now we can use the ideal gas law to calculate the pressure inside the tank. The ideal gas law is: \(PV = nRT\) Rearrange for pressure (P): \(P = \frac{nRT}{V}\) where P = pressure V = volume n = moles R = ideal gas constant (8.314 J/mol K) T = temperature in Kelvin Plug in the values for the tank: \(P = \frac{78.125\:mol * 8.314\:J/mol\:K * 283.15\:K}{11.0\:L}\) \(P = \frac{183153.2\:J/K}{11.0\:L}\) To convert J/mol to kPa, we have to divide by 100 (1 kPa = 100 J/mol): Pressure \(P = \frac{183153.2\:J/K}{11.0\:L * 100}\) \(P = 166.5\:kPa\) So, the gas pressure inside the tank is 166.5 kPa.

Step 5: (Step 5) Calculate the new volume of oxygen at given temperature and pressure

We will again use the ideal gas law to find the new volume of oxygen at \(25^{\circ}C\) and \(101.33\:kPa\). First, set up this equation: \(P_1V_1/T_1 = P_2V_2/T_2\) Rearrange to find \(V_2\): \(V_2 = \frac{P_1V_1 T_2}{P_2T_1}\) Plug in the values: \(V_2 = \frac{166.5\:kPa * 11.0\:L * 298.15\:K}{101.33\:kPa * 283.15\:K}\) \(V_2 = \frac{54617.925\:L\:K}{28683.3\: kPa\:K}\) \(V_2 = 1.90\:L\) So, the oxygen would occupy a volume of 1.90 L at \(25^{\circ}C\) and \(101.33\:kPa\).

Key Concepts

Gas Pressure Calculation

In order to calculate the gas pressure inside the tank, we use the Ideal Gas Law. This principle is fundamental in the field of chemistry and physics for understanding the behavior of gases. To find the gas pressure, we apply the formula:

• The Ideal Gas Law formula is given by: \( PV = nRT \), where:
  • \( P \) is the pressure of the gas
  • \( V \) is the volume
  • \( n \) is the number of moles
  • \( R \) is the ideal gas constant (8.314 J/mol K)
  • \( T \) is the temperature in Kelvin
• To find the pressure \( P \) inside the tank, we rearrange the equation to: \( P = \frac{nRT}{V} \).

Next, plug in the values for moles, volume, temperature, and R to calculate the specific pressure. Calculating meticulously ensures accurate results and a pressure of 166.5 kPa inside the scuba tank.

Moles of Gas

To find the number of moles of gas present in a situation like this, we use the mass-to-mole conversion. This is a necessary step for further calculations with gases as per the Ideal Gas Law. Here’s how we do it:

• First, convert the mass from kilograms to grams.
  • In our scenario, this calculation is: \( 2.50 \, \text{kg} \, \times \, 1000 \, \text{g/kg} = 2500 \, \text{g} \)
• Next, calculate the number of moles using the formula: \( \text{moles} = \frac{\text{mass of } O_2}{\text{molar mass of } O_2} \).
  • Molar mass of \( O_2 \) is approximately 32 g/mol.
  • This gives us \( \frac{2500 \, \text{g}}{32 \, \text{g/mol}} = 78.125 \, \text{mol} \).

Understanding how to calculate moles helps in predicting how gases behave under different conditions.

Temperature Conversion

Temperature conversion is crucial for calculations involving gases, particularly because the Ideal Gas Law requires temperatures to be in Kelvin. Converting Celsius to Kelvin is straightforward and can be done by simply adding 273.15 to the Celsius temperature. Here’s the process:

• Start with the temperature in Celsius.
  • For this problem, we have two temperatures to convert: \( 10^{\circ}C \) and \( 25^{\circ}C \).
• Convert each to Kelvin:
  • For \( 10^{\circ}C \): \( 10 + 273.15 = 283.15 \, \text{K} \)
  • For \( 25^{\circ}C \): \( 25 + 273.15 = 298.15 \, \text{K} \)

This conversion is essential as Kelvin is the standard unit used in gas calculations to ensure consistency and accuracy.

Volume Calculation

Calculating the volume that a gas occupies under changing conditions involves understanding both Charles's Law and Boyle's Law which are encompassed within the Ideal Gas Law. For this scenario, we use the combined gas law equation:

• First write the equation: \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \).
  • \( P_1 \) and \( V_1 \) are the initial pressure and volume; \( T_1 \) is the initial temperature in Kelvin.
  • \( P_2 \) and \( V_2 \) are the final pressure and volume; \( T_2 \) is the final temperature in Kelvin.
• Rearrange to solve for the unknown volume, \( V_2 \):
  • \( V_2 = \frac{P_1V_1T_2}{P_2T_1} \)

Plug the given values into the formula to solve for the new volume, resulting in \( 1.90\, \text{L} \) at the specified conditions of 25°C and 101.33 kPa. Accurately calculating volume changes is important in many practical situations.