Question

An empty cylindrical canister 1.50 m long and 90.0 cm in diameter is to be filled with pure oxygen at 22.0\(^\circ\)C to store in a space station. To hold as much gas as possible, the absolute pressure of the oxygen will be 21.0 atm. The molar mass of oxygen is 32.0 g/mol. (a) How many moles of oxygen does this canister hold? (b) For someone lifting this canister, by how many kilograms does this gas increase the mass to be lifted?

Short answer

(a) 828.5 moles; (b) 26.512 kg.

Step-by-step solution

Convert Measurements

First, we need to ensure all measurements are in appropriate SI units. The diameter of the canister is given in centimeters, so convert it to meters: \[ \text{Diameter} = 90.0\text{ cm} = 0.90\text{ m} \]The volume of the cylinder (canister) is given by the formula: \[ V = \pi \left( \frac{d}{2} \right)^2 h \] where \(d\) is the diameter and \(h\) is the height.

Calculate Volume

Substitute the values to find the volume: \[ V = \pi \left( \frac{0.90}{2} \right)^2 (1.50) \]Calculate the volume:\[ V \approx 0.954 \text{ m}^3 \]

Use the Ideal Gas Law

To find the number of moles, use the ideal gas law:\[ PV = nRT \]Where:- \(P\) is the pressure, \(21.0 \text{ atm} = 21.0 \times 101325 \text{ Pa} \)- \(V\) is the volume, \(0.954 \text{ m}^3\)- \(R\) is the universal gas constant, \(8.314 \text{ J/(mol} \cdot \text{K)}\)- \(T\) is the temperature in Kelvin, \(T = 22.0 + 273.15 = 295.15\text{ K}\)

Solve for Moles

Rearrange the ideal gas law equation to solve for \(n\):\[ n = \frac{PV}{RT} \]Substitute the known values:\[ n = \frac{(21.0 \times 101325) (0.954)}{8.314 \times 295.15} \]Calculate \(n\):\[ n \approx 828.5 \text{ moles} \]

Calculate Mass Increase

The molar mass of oxygen is \(32.0 \text{ g/mol}\). Convert moles to grams:\[ \text{Mass} = n \times \text{Molar Mass} = 828.5 \times 32.0 \text{ g/mol} \]Convert grams to kilograms:\[ \text{Mass} \approx 26512 \text{ g} = 26.512 \text{ kg} \]

Provide Final Answers

(a) The canister holds approximately 828.5 moles of oxygen. (b) The mass of the gas increases by about 26.512 kg.

Key concepts

Moles calculation

Calculating the number of moles of a gas involves using the Ideal Gas Law. This law relates the pressure, volume, and temperature of a gas with the number of moles. The formula is:\[ PV = nRT \]where:

• \( P \) is the pressure in pascals (Pa).
• \( V \) is the volume in cubic meters (m³).
• \( n \) is the number of moles.
• \( R \) is the ideal gas constant: 8.314 J/(mol·K).
• \( T \) is the temperature in Kelvin (K).

To solve for the number of moles, \( n \), rearrange the equation to:\[ n = \frac{PV}{RT} \]In this example, once you've converted all measurements to the correct SI units, you can substitute them into the equation and solve. This calculation shows not only the foundational relationship between these parameters but also how natural units like temperature and pressure are standardized for simplicity and computation.

Volume of a cylinder

The volume of a cylindrical object, like the canister in the problem, is calculated using the formula for the volume of a cylinder:\[ V = \pi \left( \frac{d}{2} \right)^2 h \]where:

• \( d \) is the diameter.
• \( h \) is the height (or length in this context).
• \( \pi \approx 3.14159 \)

This equation effectively accounts for the circular base area of the cylinder, \( \pi \left( \frac{d}{2} \right)^2 \), times its height. It is essential to ensure all measurements are in meters for consistency with SI units. So in the exercise, converting 90 cm to 0.9 m is crucial. When calculating, take extra care to handle the division and power correctly.

Pressure conversion

Pressure conversion is a vital step in gas law calculations as pressure might not be initially in the required units. The most common unit used in computations is pascals (Pa). In our example, the provided pressure is in atmospheres (atm), and we convert it to pascals using the conversion factor:\[ 1 \, \text{atm} = 101325 \, \text{Pa} \]This conversion is straightforward: just multiply the atmospheric pressure by 101325. Thus, 21.0 atm becomes:\[ 21.0 \times 101325 \, \text{Pa} \]Converting to pascals ensures compatibility with the Ideal Gas Law, as the constant \( R \) is in terms of energy per mole per Kelvin, requiring pressure in pascals to balance the equation dimensions.

Temperature conversion

In scientific calculations involving gases, temperature must always be in Kelvin. The Kelvin scale is an absolute temperature scale, where 0 K is the lowest theoretical temperature. To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature:\[ T (K) = T (°C) + 273.15 \]For the exercise, a temperature of 22.0 °C is converted to Kelvin:\[ 22.0 + 273.15 = 295.15 \, K \]This conversion is crucial because the Kelvin scale directly corresponds to kinetic energy and ensures the Ideal Gas Law calculations reflect the true behavior of gas particles. Maintaining temperature in Kelvin allows consistent use of the ideal gas constant \( R \) across different problems.