Question

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is \(65.0 \mathrm{~L}\) and which 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 \(55.0 \mathrm{~L} ?\)

Short answer

a) The mass of O₂ in the tank is 7559 g. b) The volume of O₂ at STP is 5,053 L. c) The temperature at which the pressure in the tank equals 150.0 atm is 158°C. d) The pressure of the gas, in kPa, when transferred to a container at 24°C with a volume of 55.0 L, is 22,854 kPa.

Step-by-step solution

a) Mass of O₂ in the tank

First, convert the given temperature and pressure to the appropriate units for the ideal gas law: Temperature: 23°C + 273 = 296 K Pressure: \( 16,500 kPa * \frac{1 atm}{101.3 kPa} \approx 162.9 atm \) Now we can use the ideal gas law: PV = nRT Solving for n (number of moles), we have \( n = \frac{PV}{RT} \) \( n = \frac{(162.9 atm)(65.0 L)}{(0.0821 \frac{L \cdot atm}{mol \cdot K})(296 K)} \approx 236.2 mol \) The molar mass of oxygen is 32 g/mol, so the mass of O₂ in the tank is: \( m = n * molar ~mass = 236.2 mol * 32 \frac{g}{mol} = 7559 g \)

b) Volume of O₂ at STP

Standard Temperature and Pressure (STP) are defined as 0°C (or 273 K) and 1 atm. Using the ideal gas law and the n value we found in part (a), we can find the volume at STP: \( V = \frac{nRT}{P} \) \( V = \frac{(236.2 mol)(0.0821 \frac{L \cdot atm}{mol \cdot K})(273 K)}{1 atm} \approx 5,053 L \)

c) Temperature for a pressure of 150.0 atm

Given the pressure, we need to find the temperature at which this occurs. Rearrange the ideal gas law and solve for temperature: \( T = \frac{PV}{nR} \) \( T = \frac{(150 atm)(65.0 L)}{(236.2 mol)(0.0821 \frac{L \cdot atm}{mol \cdot K})} \approx 431 K \) Convert this temperature back to Celsius: \( 431 K - 273 = 158^{\circ}C \)

d) Pressure at 24°C and in a 55.0 L container

Convert the new temperature to Kelvin: 24°C + 273 = 297 K Now we can use the ideal gas law to find the pressure of the gas in these new conditions: \( P = \frac{nRT}{V} \) \( P = \frac{(236.2 mol)(0.0821 \frac{L \cdot atm}{mol \cdot K})(297 K)}{55.0 L} \approx 225.7 atm \) Finally, convert the pressure to kPa: \( 225.7 atm * 101.3 \frac{kPa}{atm} \approx 22,854 kPa \) The answers are: a) 7559 g b) 5,053 L c) 158°C d) 22,854 kPa

Key concepts

Gas Pressure and Volume

The relationship between gas pressure and volume is intriguing and forms a fundamental aspect of the behavior of gases. Underlying this is the principle that gases consist of tiny particles in constant, random motion. When these particles collide with the walls of a container, they exert a force, and that force per unit area is what we refer to as gas pressure. Now, imagine you have a container with a fixed amount of gas at a constant temperature.

STP Conditions

In the realm of chemistry and physics, STP stands for Standard Temperature and Pressure. It serves as a reference point, allowing scientists and students to compare different sets of gas data. At STP, the standard temperature is precisely 0°C or 273.15K, and the standard pressure is 1 atmosphere (atm). These conditions are vital when using the ideal gas law because the volume that a gas occupies can drastically change with temperature and pressure.

Molar Mass of Gases

The molar mass of a gas is a measure of the mass of one mole of that gas. It is a critical component when dealing with gas calculations because it allows us to connect the mass of a gas to the number of moles. Oxygen, for instance, has a molar mass of approximately 32 grams per mole (g/mol). Calculating the mass of a gas from its volume, or vice versa, involves using the ideal gas law in conjunction with the molar mass.

Converting Units in Gas Calculations

Converting units is a common practice in gas calculations to maintain consistency and accuracy in the results. Since the ideal gas law includes pressure (P), volume (V), temperature (T), and number of moles (n), it is crucial to use appropriate units for each. Usually, pressure is converted to atmospheres (atm) or kilopascals (kPa), volume to liters (L), temperature to Kelvin (K), and n remains in moles (mol). This ensures that when these measurements are put into the ideal gas law equation, they are compatible with the gas constant (R).