Problem 39 - Chemistry The Central Science In Si Units

Question

A scuba diver's tank contains \(2.50 \mathrm{~kg}\) of \(\mathrm{O}_{2}\) compressed into a volume of 11.0 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

(a) The pressure is 1664.79 kPa. (b) The volume is 1924.14 L.

Step 1: Calculate Moles of Oxygen

To find the moles of oxygen, use its molar mass. The molar mass of \ \( \mathrm{O}_2 \ \) is \ \(32.00 \, \mathrm{g/mol} \). Convert the mass from kilograms to grams: \ \(2.50 \, \mathrm{kg} = 2500 \, \mathrm{g} \). Then use the formula \ \(n = \frac{\text{mass}}{\text{molar mass}} = \frac{2500 \, \mathrm{g}}{32.00 \, \mathrm{g/mol}} = 78.125 \, \text{mol} \).

Step 2: Convert Temperature from Celsius to Kelvin for Part A

The temperature must be in Kelvin for gas calculations. Convert \ \(10^{\circ} \mathrm{C} \) to Kelvin: \ \(T = 10 + 273.15 = 283.15 \, \mathrm{K} \).

Step 3: Calculate Pressure Using the Ideal Gas Law for Part A

Use the ideal gas law \ \(PV = nRT \), where \ \(R = 8.314 \, \mathrm{J/(mol \cdot K)} \), converted to \ \( \, \text{L} \cdot \text{kPa/mol} \cdot \text{K} \): \ \(R = 8.314 \times 10^{-3} \). Rearrange to solve for \ \(P \:\: P = \frac{nRT}{V} = \frac{78.125 \, \mathrm{mol} \cdot 8.314 \, \text{J/mol K} \cdot 283.15}{11.0 \, \text{L}} = 1664.79 \, \mathrm{kPa} \).

Step 4: Convert Temperature from Celsius to Kelvin for Part B

For the second part, convert \ \(25^{\circ} \mathrm{C} \) to Kelvin: \ \(T = 25 + 273.15 = 298.15 \, \mathrm{K} \).

Step 5: Calculate New Volume Using the Ideal Gas Law for Part B

To find the new volume, rearrange the ideal gas law to \ \(V = \frac{nRT}{P} \:\: V = \frac{78.125 \, \text{mol} \cdot 8.314 \, \text{J/mol K} \cdot 298.15}{101.33} \: V = 1924.14 \, \mathrm{L} \).

Key Concepts

Gas Pressure Calculation

The calculation of gas pressure inside a container is crucial for understanding various gas behaviors, especially when diving. The Ideal Gas Law, expressed as \( PV = nRT \), can be used to calculate the pressure within a tank. In the formula:

• \( P \) represents the pressure,
• \( V \) is the volume of the gas,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant,
• \( T \) is the absolute temperature in Kelvin.

To find the gas pressure, rearrange the formula to solve for \( P \): \[ P = \frac{nRT}{V} \] Given our example with 78.125 moles of \( \mathrm{O}_2 \), a gas constant \( R \) of 8.314, a temperature of 283.15 K, and a volume of 11.0 L, the pressure is calculated to be 1664.79 kPa. Remember, the accuracy of this calculation depends on using consistent units for R and ensuring the temperature is in Kelvin.

Temperature Conversion

Temperature conversion is an essential part of gas calculations, as the Ideal Gas Law requires temperature to be in Kelvin. To convert Celsius to Kelvin, we use the simple formula: \[ T(\mathrm{K}) = T(^{\circ}C) + 273.15 \] For part (a) of the problem, we converted \( 10^{\circ} \mathrm{C} \) to Kelvin, resulting in 283.15 K. For part (b), \( 25^{\circ} \mathrm{C} \) becomes 298.15 K. This conversion ensures that all inputs are compatible with the Ideal Gas Law, preventing errors in pressure or volume calculations.

Moles of Gas

Understanding the concept of moles is vital in chemistry as it provides a bridge between the mass of a substance and the number of particles or molecules. To find the number of moles \( n \), we divide the mass of the substance by its molar mass: \[ n = \frac{\text{mass}}{\text{molar mass}} \] In this example, the tank contains 2.50 kg of \( \mathrm{O}_2 \). First, convert 2.50 kg to grams by multiplying by 1000, yielding 2500 g. Then, using the molar mass of \( \mathrm{O}_2 \), which is 32.00 g/mol, calculate the moles: \[ n = \frac{2500 \, \mathrm{g}}{32.00 \, \mathrm{g/mol}} = 78.125 \, \text{mol} \] This gives us the amount of substance needed for the Ideal Gas Law in gas behavior calculations.

Volume Calculation

Calculating the volume a gas would occupy at different temperatures and pressures involves manipulating the Ideal Gas Law. Once again, the equation \( PV = nRT \) comes into play. Rearrange it to find volume: \[ V = \frac{nRT}{P} \] In this scenario, we want to know the volume where the temperature is 298.15 K and the pressure is 101.33 kPa. Given the same 78.125 moles and using the appropriate gas constant \( R \), the volume \( V \) is calculated as: \[ V = \frac{78.125 \, \text{mol} \times 8.314 \, \text{J/mol K} \times 298.15 \mathrm{K}}{101.33}= 1924.14 \, \mathrm{L} \] Hence, this calculation indicates the space the gas would occupy under those conditions, illustrating how environmental conditions like temperature and pressure critically influence gas behavior.