Question

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is 55.0 gallons that contains \(\mathrm{O}_{2}\) gas at a pressure of \(16,500 \mathrm{kPa}\) at \(23{ }^{\circ} \mathrm{C}\). (a) What mass of \(\mathrm{O}_{2}\) does the tank contain? (b) What volume would the gas occupy at STP? (c) At what temperature would the pressure in the tank equal 150.0 atm? (d) What would be the pressure of the gas, in \(\mathrm{kPa},\) if it were transferred to a container at \(24^{\circ} \mathrm{C}\) whose volume is \(55.0 \mathrm{~L} ?\)

Short answer

The mass of \(\mathrm{O}_{2}\) in the tank is approximately 4662.17 g. At STP, the gas would occupy a volume of approximately 3267.93 L. The temperature at which the pressure in the tank equals 150.0 atm is approximately 324.60 K. The pressure of the gas in the new container is approximately 75207.33 kPa.

Step-by-step solution

Apply the Ideal Gas Law to find moles of \(\mathrm{O}_{2}\)

Using the Ideal Gas Law formula, \(PV=nRT\), solve for the number of moles (n) in the tank: \(n = \frac{PV}{RT}\) We can then plug in our given values: n = \(\frac{162.861 \text{ atm} \cdot 208.197 \text{ L}}{(0.0821(\text{L}\cdot\text{atm})/(\text{mol}\cdot\text{K})) \cdot 296.15 \text{ K}}\) n ≈ 145.818 mol

Calculate the mass of \(\mathrm{O}_{2}\)

We know the number of moles (n) and the molar mass of \(\mathrm{O}_{2}\). Now, the mass (m) can be calculated using the formula: mass of \(\mathrm{O}_{2}\) (m) = n × molar mass of \(\mathrm{O}_{2}\) m = 145.818 mol × 32.00 g/mol m ≈ 4662.17 g The mass of \(\mathrm{O}_{2}\) in the tank is approximately 4662.17 g. b) Volume the gas would occupy at STP (STP: standard temperature and pressure; P=1 atm, T=273.15 K)

Calculate the volume at STP

Using the Ideal Gas Law formula, \(PV=nRT\), solve for the volume (V) at STP: V = \(\frac{nRT}{P}\) We can then plug in the values at STP: V = \(\frac{145.818 \text{ mol} \cdot (0.0821(\text{L}\cdot\text{atm})/(\text{mol}\cdot\text{K})) \cdot 273.15 \text{ K}}{1 \text{ atm}}\) V ≈ 3267.93 L At STP, the gas would occupy a volume of approximately 3267.93 L. c) Temperature at which the pressure in the tank equals 150.0 atm

Calculate the temperature

Using the Ideal Gas Law formula, \(PV=nRT\), solve for the temperature (T) when pressure is 150.0 atm: T = \(\frac{PV}{nR}\) We can then plug in the values: T = \(\frac{150.0 \text{ atm} \cdot 208.197 \text{ L}}{145.818 \text{ mol} \cdot (0.0821(\text{L}\cdot\text{atm})/(\text{mol}\cdot\text{K}))}\) T ≈ 324.60 K The temperature at which the pressure in the tank equals 150.0 atm is approximately 324.60 K. d) Pressure of the gas in a 55.0 L container at 24°C Temperature at 24°C = 297.15 K

Calculate the new pressure

Using the Ideal Gas Law formula, \(PV=nRT\), solve for the pressure (P) in the new container: P = \(\frac{nRT}{V}\) We can then plug in the values: P = \(\frac{145.818 \text{ mol} \cdot (0.0821(\text{L}\cdot\text{atm})/(\text{mol}\cdot\text{K})) \cdot 297.15 \text{ K}}{55.0 \text{ L}}\) P ≈ 742.431 atm Now, let's convert this pressure to kPa: P = 742.431 atm × 101.325 kPa/atm P ≈ 75207.33 kPa The pressure of the gas in the new container is approximately 75207.33 kPa.

Key concepts

Molar Mass

Molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a substance.
For the exercise involving oxygen gas, knowing the molar mass of oxygen (\( \mathrm{O}_{2} \)) is crucial to determine how much mass is present within a mole.
- **Molecular Weight of \( \mathrm{O}_{2} \):** Oxygen has an atomic weight of approximately 16.00 g/mol. Since \( \mathrm{O}_{2} \) is diatomic, we calculate its molar mass by multiplying the atomic weight by two. Therefore, the molar mass of \( \mathrm{O}_{2} \) is \( 32.00 \text{ g/mol} \).
- **Application in Calculations:** When we know the number of moles of \( \mathrm{O}_{2} \) in a container, we can easily calculate the mass by multiplying the number of moles (\( n \)) by the molar mass (\( M \)) using the formula: \[m = n \times M\]This is essential for determining how much oxygen is stored, which was approximately 4662.17 g in the original problem.

Standard Temperature and Pressure (STP)

Standard Temperature and Pressure (STP) is a reference point used in chemistry to allow scientists to make consistent measurements.
- **Conditions at STP:** STP is typically set at a temperature of \( 0^{\circ}\text{C} \) or 273.15 Kelvin (K) and a pressure of \( 1 \text{ atm} \).
These conditions are commonly used as they provide a convenient baseline for comparing gas behaviors.
- **Ideal Gas Law Application:** Using STP conditions, you can calculate the volume a gas would occupy at these standardized settings using the Ideal Gas Law. For example, in the exercise solution, the computation at STP reveals that the gas would occupy a volume of about 3267.93 L. This allows for easy comparison and understanding of how gas volumes change under different conditions.

Temperature Conversion

Understanding temperature conversions is vital, especially when dealing with gas laws that require temperatures to be in Kelvin.- **Why Kelvin?:** The Kelvin scale is used in scientific calculations because it begins at absolute zero, where theoretically, particles possess minimum thermal motion.
Converting Celsius to Kelvin is simple: add 273.15 to the Celsius temperature. For instance, in the problem's step-by-step solution:\[\text{Temperature in Kelvin} = \text{Temperature in Celsius} + 273.15\]So, \( 23^{\circ}\text{C} \) converts to \( 296.15 \text{ K} \), and \( 24^{\circ}\text{C} \) becomes \( 297.15 \text{ K} \).- **Importance in the Ideal Gas Law:** Working in Kelvin is necessary since the Gas Constant (\( R \)) used in the Ideal Gas Law is typically given in units of \( L\cdot atm/mol\cdot K \). Correct temperature conversion ensures accurate calculations of pressure, volume, and moles of gas.