Problem 44 - Chemistry: The Central Science in SI Units, Global Edition

Question

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is \(210.0 \mathrm{~L}\) 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 \(15.2 \mathrm{MPa} ?\) (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 \(55.0 \mathrm{~L} ?\)

Step-by-Step Solution

Answer

In summary, a steel tank with a volume of 210 L containing O2 gas under the given conditions has a mass of approximately 202.56 g. At STP, the gas would occupy a volume of 137.1 L. The temperature at which the pressure would equal 15.2 MPa is approximately 452.47 K. If the gas were transferred to a 55.0 L container at 24°C, the pressure would be approximately 51,099 kPa.

Step 1: Part a: Calculate the mass of the O2 gas in the tank

First, we need to convert the provided information into appropriate units. The tank volume is given in Liters, the pressure is given in kPa, and the temperature is given in Celsius. The values we need are: \(V = 210.0 L\) \(P = 16,500 kPa = 16,500,000 Pa\) (since 1kPa = 1000Pa) \(T = 23^{\circ} C = 296 K\) (since T(K) = T(°C) + 273) Now, we can use the ideal gas law formula to solve for the moles of O2 gas: \(PV = nRT\) \(n = \frac{PV}{RT}\) \(n = \frac{16,500,000 Pa\times210 L}{8.314 J/(mol K)\times296 K}\) \(n = 15526.7/2454\) \(n \approx 6.33 mol\) Finally, to find the mass of O2, we multiply the moles by the molar mass of O2: Mass = moles × molar mass of O2 Mass = \(6.33 mol \times 32 g/mol\) Mass = \(202.56 g\) Therefore, the steel tank contains approximately 202.56 g of O2.

Step 2: Part b: Calculate the volume of the gas at STP

STP (standard temperature and pressure) is defined as a temperature of 273 K and a pressure of 100 kPa. To find the volume the gas would occupy at STP, we can use the ideal gas law formula again. This time we need to consider the number of moles and the constant for ideal gases given as the constant for the ideal gases in \(L\cdot kPa/(mol\cdot K)\), \(R=8.31 L\cdot kPa/(mol\cdot K)\): \(V = \frac{nRT}{P}\) \(V = \frac{6.33 mol\times8.31 L\cdot kPa/(mol\cdot K)\times273 K}{100 kPa}\) \(V \approx 137.10 L\) So, the volume of the gas at STP would be approximately 137.1 L.

Step 3: Part c: Calculate the temperature at which the pressure equals 15.2 MPa

We have been given the pressure as 15.2 MPa, which is equal to 15,200 kPa. We will use the ideal gas law formula to find the temperature at this particular pressure, considering the number of moles, volume, and ideal gas constant given: \(T = \frac{PV}{nR}\) \(T = \frac{15,200 kPa\times210 L}{6.33 mol\times8.31 L\cdot kPa/(mol\cdot K)}\) \(T \approx 452.47 K\) The temperature at which the pressure in the tank equals 15.2 MPa is approximately 452.47 K.

Step 4: Part d: Calculate the pressure at 24°C with a 55.0 L container volume

The volume of the container is given as 55.0 L and the temperature is given in Celsius as 24°C which corresponds to 297 K. We can use the ideal gas law formula to find the pressure under these conditions: \(P = \frac{nRT}{V}\) \(P = \frac{6.33 mol\times8.31 L\cdot kPa/(mol\cdot K)\times297 K}{55.0 L}\) \(P \approx 51,099 kPa\) The pressure of the gas in a 55.0 L container at 24°C would be approximately 51,099 kPa.

Key Concepts

Gas Pressure Calculations

To understand how gas pressure calculations are made, it's crucial to start with the Ideal Gas Law, an essential equation in chemistry. The Ideal Gas Law is expressed as:

• \(PV = nRT\)

Where:

• \(P\) is the pressure of the gas in pascals (Pa).
• \(V\) is the volume in liters (L).
• \(n\) is the number of moles of the gas.
• \(R\) is the ideal gas constant \(8.314\, \text{J/(mol K)}\) or \(0.0821\, \text{L atm/(mol K)}\).
• \(T\) is the temperature in Kelvin (K).

When dealing with gas under varying conditions, like in our problem, we need to adjust for the unit conversions. For example, converting pressure units from kPa to Pa or temperature from Celsius to Kelvin by adding 273 to the Celsius value. Breaking down calculations into smaller, manageable steps using these conversions is imperative for accuracy.In the given problem, adjustments for the pressure and temperature units were made to apply the Ideal Gas Law effectively.Once all units are uniform, you can calculate other properties like moles or mass easily using calculations derived from this formula.

Standard Temperature and Pressure

Standard Temperature and Pressure (STP) is a baseline set of conditions in chemistry that helps standardize calculations and comparisons across different gases. STP is defined by a temperature of 273 K (0°C) and a pressure of 100 kPa.
Understanding STP is vital when calculating gas volumes or engaging in any gas law problems, as many problems reference these conditions explicitly.In the context of this exercise, determining the volume of the gas at STP involves recalculating the gas' volume under standard conditions while keeping the moles constant.By using the equation:

• \(V = \frac{nRT}{P}\)

the concept helps simplify comparisons and provides a clear uniform point of reference, making it easier to determine whether gases behave ideally under identical conditions or if there are deviations to consider due to real-life constraints.

Molar Mass Calculation

Calculating molar mass is a key step in solving gas law problems. Molar mass connects the number of moles to the mass of a substance in grams. The molar mass of a compound is calculated by summing the atomic masses of the constituent elements.
For oxygen gas (\(\text{O}_2\)), the molar mass is found by adding the atomic masses of two oxygen atoms:

• \(16 \text{ g/mol per oxygen atom} \times 2 = 32 \text{ g/mol for } \text{O}_2\)

Once the number of moles is known, the mass can be determined by the formula:

• \(\text{Mass} = \text{moles} \times \text{molar mass}\)

This relationship is employed in the original exercise to find the mass of oxygen in the tank. By knowing the moles obtained from the Ideal Gas Law and the molar mass, you can efficiently calculate the total mass of the gas present. Thus, understanding molar mass is critical for going from theoretical calculations to practical applications in laboratory or industrial situations.