Problem 44 - Chemistry: The Central Science

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 \(150.0 \mathrm{~atm}\) ? (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 \(55.0 \mathrm{~L}\) ?

Step-by-Step Solution

Answer

(a) The tank contains approximately 494,000 grams of O₂. (b) The gas would occupy a volume of approximately 350.49 \(m^3\) at STP. (c) The temperature at which the pressure in the tank equals 150 atm is approximately 301.57 K. (d) The pressure of the gas in the new container is approximately 6,611 kPa.

Step 1: Convert quantities to SI units

First, we need to convert the given values to SI units: Volume: 55.0 gallons ⟶ cubic meters (\(m^3\)) Pressure: 16,500 kPa ⟶ Pascals (Pa) Temperature: 23 °C ⟶ Kelvin (K) Using the following conversions: 1 gallon = 0.00378541 \(m^3\) 1 kPa = 1000 Pa K = °C + 273.15 We get: Volume (V) = 55 × 0.00378541 \(m^3\) ≈ 0.208 L Pressure (P) = 16,500 × 1,000 Pa = 16,500,000 Pa Temperature (T) = 23 + 273.15 = 296.15 K

Step 2: (a) Calculate the number of moles and mass of O₂

We are given the pressure (P), volume (V), and temperature (T). We can use the Ideal Gas Law to find the number of moles (n) by rearranging the formula: \(n = \frac{PV}{RT}\). Using the gas constant R = 8.314 J/(mol·K), we get: \(n = \frac{16,500,000 \times 0.208}{8.314 \times 296.15} \approx 15444\) moles To find the mass (m) of O₂, we'll use the molecular weight of oxygen gas, which is 32 g/mol: m = n × Molecular weight = 15444 × 32g/mol ≈ 494,000 g (a) The tank contains approximately 494,000 grams of O₂.

Step 3: (b) Calculate the volume at STP

Standard Temperature and Pressure (STP) are defined as 0 °C (273.15 K) and 100 kPa (100,000 Pa) respectively. To find the volume (V) at STP, we can use the Ideal Gas Law again: \(V = \frac{nRT}{P}\). Using the standard temperature (T = 273.15 K) and pressure (P = 100,000 Pa), we get: \(V = \frac{15444 \times 8.314 \times 273.15}{100,000} \approx 350.49 m^3\) (b) The gas would occupy a volume of approximately 350.49 \(m^3\) at STP.

Step 4: (c) Find the temperature at 150 atm

We need to find the temperature (T) when the pressure (P) is 150 atmospheres (atm). Convert 150 atm to Pascals (Pa) using the conversion 1 atm = 101325 Pa: P = 150 × 101,325 Pa = 15,198,750 Pa Now, we can rearrange the Ideal Gas Law formula to find the temperature: \(T = \frac{PV}{nR}\). \(T = \frac{15,198,750 \times 0.208}{15444 \times 8.314} \approx 301.57 K\) (c) The temperature at which the pressure in the tank equals 150 atm is approximately 301.57 K.

Step 5: (d) Calculate the new pressure after transferring to a new container

Finally, find the pressure (P) in the new container at 24 °C (297.15 K) with a volume of 55.0 L. Convert the volume to \(m^3\): V = 55.0 L × 0.001 = 0.055 \(m^3\) Now, use the Ideal Gas Law formula to find the pressure: \(P = \frac{nRT}{V}\). \(P = \frac{15444 \times 8.314 \times 297.15}{0.055} \approx 6,611,100 Pa\) Convert the pressure to kPa: P = 6,611,100 Pa × 0.001 ≈ 6,611 kPa (d) The pressure of the gas in the new container is approximately 6,611 kPa.

Key Concepts

Gas Laws

The Ideal Gas Law is a cornerstone of chemistry and physics when dealing with gases. It is represented by the formula \( PV = nRT \), where \( P \) stands for pressure, \( V \) is the volume, \( n \) denotes the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. This law offers a simple way to relate these properties of any gas under the assumption of ideal conditions.

The Ideal Gas Law is pivotal because it can be utilized to determine any one of the gas's properties if the others are known. It works under the premise that gas molecules are minute and non-interacting, which closely approximates real gases unless under extreme conditions like very high pressure or low temperature.

By examining exercises like this, students learn to apply the ideal gas equation to different scenarios such as calculating the number of moles or altering the system’s conditions to find associated impact on pressure or volume.

Pressure Conversion

Pressure is often given in various units depending on the context, such as kilopascals (kPa), atmospheres (atm), or Pascals (Pa). Converting between these units is a vital skill. For instance, in this exercise, we needed to convert 16,500 kPa to Pascals, since the ideal gas constant \( R \) is in Joules per mole per Kelvin, necessitating consistent units. Knowing that 1 kPa is equivalent to 1,000 Pa, we simply multiply by 1,000 to get everything to Pa.

Similarly, when converting from atmospheres, we use the fact that 1 atm equals 101,325 Pa. So, 150 atm becomes 15,198,750 Pa when converted. Understanding these conversions is essential for solving problems using the ideal gas law because consistency in units is crucial.

Temperature Conversion

Temperature must always be converted to Kelvin (K) when using the Ideal Gas Law. This is because the Kelvin scale is an absolute temperature scale, starting from absolute zero, and is directly proportional to the energy of the particles.

To convert from degrees Celsius to Kelvin, simply add 273.15. In the exercise, the room temperature of 23°C was converted to 296.15 K, and 24°C became 297.15 K. This simple addition ensures that all temperatures are in the correct scale for the Ideal Gas Law calculations.

Converting temperature correctly is crucial as it influences the other parameters significantly, and errors can lead to incorrect results due to the direct proportionality in the Ideal Gas Law.

Volume Conversion

Volume conversion becomes necessary because various units like liters (L), gallons, and cubic meters (m³) are used in different contexts. The ideal gas constant \( R \) uses m³, making it essential to convert any given volume into this unit.

In the exercise, we converted 55.0 gallons to cubic meters using the conversion factor 1 gallon = 0.00378541 m³, resulting in approximately 0.208 m³. Similarly, a new volume of 55.0 L was converted to 0.055 m³ by multiplying with 0.001.

Understanding and being able to convert these units is key when engaging with gas laws to assure all data is compatible, leading to accurate calculations and insights.