Question

A scuba diver's tank contains 0.29 \(\mathrm{kg}\) of \(\mathrm{O}_{2}\) compressed into a volume of 2.3 \(\mathrm{L}\) . (a) Calculate the gas pressure inside the tank at \(9^{\circ} \mathrm{C} (\mathbf{b})\) What volume would this oxygen occupy at \(26^{\circ} \mathrm{C}\) and 0.95 atm?

Short answer

The gas pressure inside the scuba diver's tank is approximately \(9.47\cdot10^{5}\:Pa\). The volume that the oxygen occupies at 26°C and 0.95 atm is approximately \(26.42\:L\).

Step-by-step solution

Convert the given temperature to Kelvin

To use the ideal gas law equation, we need to convert the given temperatures to Kelvin. To do this, add 273.15 to the given temperatures in degrees Celsius: - For 9°C: \(T_1 = 9 + 273.15 = 282.15\:K\) - For 26°C: \(T_2 = 26 + 273.15 = 299.15\:K\)

Determine the number of moles of oxygen

Given the mass of oxygen, \(0.29\:kg\), we can calculate the number of moles (n) using the molar mass of oxygen (\(M_{O2} = 32\:g/mol\)). First, convert the mass of oxygen into grams: 0.29 kg = 290 g Next, use the molar mass to determine the number of moles of oxygen: \[n = \frac{mass}{molar\:mass} = \frac{290}{32} = 9.0625\:moles\]

Calculate the gas pressure inside the tank (part a)

Use the ideal gas law equation, \(PV = nRT\), to find the pressure (P) inside the tank. Rearrange the equation to solve for P: \[P = \frac{nRT}{V}\] We know the values of n (\(9.0625\:moles\)), R (\(8.314\:J/(mol\:K)\)), T (\(282.15\:K\)), and V (\(0.0023\:m^{3}\, since 1 L = 0.001 m^{3}\)). Substitute these values into the equation: \[P = \frac{(9.0625)(8.314)(282.15)}{0.0023} = 947014.95\:Pa\] So, the gas pressure inside the tank is approximately \(9.47\cdot10^{5}\:Pa\).

Calculate the volume the oxygen occupies at different conditions (part b)

To find the volume the oxygen occupies at different conditions, we again use the ideal gas law equation. This time, we have the initial pressure (P1), initial volume (V1), initial temperature (T1), final pressure (P2), final temperature (T2), and we need to find the final volume (V2). First, convert the pressure to atm: \[P_1 = 9.47\cdot10^{5} / 101325 = 9.35\:atm\] Now, use the combined gas law equation: \[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\] Rearrange the equation to solve for \(V_2\): \[V_2 = \frac{P_1V_1T_2}{P_2T_1}\] Substitute the values into the equation: \[V_2 = \frac{(9.35\:atm)(2.3\:L)(299.15\:K)}{(0.95\:atm)(282.15\:K)} = 26.42\:L\] So, the volume that the oxygen occupies at 26°C and 0.95 atm is approximately \(26.42\:L\).

Key concepts

Gas Pressure Calculation

Understanding gas pressure calculation is essential in many scientific applications, including diving, where accurate pressure measurements ensure safety and proper equipment function. Gas pressure is defined as the force that a gas exerts on the walls of its container, and it is directly related to the number of gas particles in a given volume and the temperature of the gas.

To calculate the pressure (P) of a gas, the ideal gas law equation is typically used:
\[P = \frac{nRT}{V}\]
Where:

• \(n\) is the number of moles of gas,
• \(R\) is the universal gas constant, 8.314 J/(mol·K),
• \(T\) is the absolute temperature in Kelvin (K),
• \(V\) is the volume in cubic meters.

By rearranging and substituting the known variables into this equation, we can solve for the unknown pressure of a gas. For instance, in the scenario with the scuba diver's oxygen tank, the pressure calculation revealed a pressure of approximately 947,015 Pascals after the values were plugged into the equation.

Understanding how pressure changes under different conditions is critical, not only for scuba divers but also for chemists, physicists, and engineers who work with gases in various contexts.

Temperature Conversion Kelvin

Temperature plays a vital role in calculating gas pressure and volume as it impacts the energy of gas particles. The Ideal Gas Law requires the temperature to be in Kelvin (K), the base unit for thermodynamic temperature measurement in the International System of Units (SI).

Converting Celsius to Kelvin is a straightforward process:
\[K = \degree C + 273.15\]
For example, to convert 9°C to Kelvin, one would perform the following calculation:
\[T_1 = 9 + 273.15 = 282.15 K\]
This process is critical for correct pressure and volume calculations using the ideal gas law. Failing to convert to Kelvin could result in inaccurate results and, in practical applications, could pose safety risks.

Moles of Gas

In chemistry, the 'mole' is a fundamental concept that represents a specific number of particles. One mole of any substance corresponds to approximately 6.022 x 10²³ particles, known as Avogadro's number.

Calculating moles of gas is essential for understanding the quantity of gas involved in reactions or contained in vessels. To find the number of moles (\(n\)), one can use the formula:
\[n = \frac{mass}{molar\:mass}\]

• The mass must be in grams (g).
• The molar mass is specific to each substance and is measured in grams per mole (g/mol).

In our textbook case, converting the mass of oxygen to moles was crucial to calculate gas pressure, since the ideal gas law relates the number of moles to pressure, volume, and temperature.

Combined Gas Law

The combined gas law integrates Boyle’s Law, Charles's Law, and Gay-Lussac’s Law, providing a comprehensive equation for situations where pressure, volume, and temperature of a gas change simultaneously. The combined gas law formula is:
\[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\]
This equation allows you to calculate the unknown state of a gas (\(P_2\), \(V_2\), or \(T_2\)), provided that the other variables are known from the initial state of the gas. Here, \(P_1\), \(V_1\), and \(T_1\) correspond to the initial pressure, volume, and temperature, and the subscripts '2' refer to the subsequent or final conditions.

In the scuba tank scenario from the textbook, by knowing the initial and final conditions of temperature and pressure, we applied the combined gas law to calculate the new volume of gas at different conditions. The transition of oxygen volume from 2.3 liters at initial conditions to approximately 26.42 liters at the specified final conditions reflects a practical example of this law at work.