Question

A welder using a tank of volume 0.0750 m\(^3\) fills it with oxygen (molar mass 32.0 g/mol) at a gauge pressure of 3.00 \(\times\) 10\({^5}\) Pa and temperature of 37.0\(^\circ\)C. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is 22.0\(^\circ\)C, the gauge pressure of the oxygen in the tank is 1.80 \(\times\) 10\({^5}\) Pa. Find (a) the initial mass of oxygen and (b) the mass of oxygen that has leaked out.

Short answer

Initial mass: 372.16 g; Leaked mass: 99.52 g.

Step-by-step solution

Convert Temperatures

The problem provides temperatures in Celsius, which must be converted to Kelvin for calculations. Convert the initial temperature of 37.0°C and final temperature of 22.0°C to Kelvin using the formula: \[ T(K) = T(°C) + 273.15 \] Thus, the initial temperature is 310.15 K and the final temperature is 295.15 K.

Calculate Initial Moles of Oxygen

Using the ideal gas law \( PV = nRT \), where \( R = 8.314 \) J/mol·K, solve for the initial number of moles \( n \). The initial pressure \( P \) is gauge pressure plus atmospheric pressure \((1.00 \times 10^5\) Pa):\[ P_{initial} = 3.00 \times 10^5 \, \text{Pa} + 1.00 \times 10^5 \, \text{Pa} = 4.00 \times 10^5 \, \text{Pa} \]Substitute the values:\[ n_{initial} = \frac{P_{initial} \times V}{R \times T_{initial}} = \frac{4.00 \times 10^5 \, \text{Pa} \times 0.0750 \, \text{m}^3}{8.314 \, \text{J/mol·K} \times 310.15 \, \text{K}} \approx 11.63 \, \text{moles} \]

Calculate Initial Mass of Oxygen

Using the molar mass of oxygen (32.0 g/mol), calculate the initial mass:\[ \text{Mass}_{initial} = n_{initial} \times \text{molar mass} = 11.63 \, \text{moles} \times 32.0 \, \text{g/mol} \approx 372.16 \, \text{g} \]

Calculate Final Moles of Oxygen

Now, calculate the final number of moles after some oxygen has leaked, using the final conditions:\[ P_{final} = 1.80 \times 10^5 \, \text{Pa} + 1.00 \times 10^5 \, \text{Pa} = 2.80 \times 10^5 \, \text{Pa} \]\[ n_{final} = \frac{P_{final} \times V}{R \times T_{final}} = \frac{2.80 \times 10^5 \, \text{Pa} \times 0.0750 \, \text{m}^3}{8.314 \, \text{J/mol·K} \times 295.15 \, \text{K}} \approx 8.52 \, \text{moles} \]

Calculate Final Mass of Oxygen

Calculate the final mass using the molar mass:\[ \text{Mass}_{final} = n_{final} \times \text{molar mass} = 8.52 \, \text{moles} \times 32.0 \, \text{g/mol} \approx 272.64 \, \text{g} \]

Calculate Mass of Oxygen Leaked

Determine the mass of oxygen that has leaked out by subtracting the final mass from the initial mass:\[ \text{Mass}_{leaked} = \text{Mass}_{initial} - \text{Mass}_{final} = 372.16 \, \text{g} - 272.64 \, \text{g} \approx 99.52 \, \text{g} \]

Key concepts

Molar Mass

Molar mass is a fundamental property of a substance that tells us about the mass of one mole of its particles. For substances like gases, understanding molar mass is crucial to making sense of various calculations, especially those involving the ideal gas law.
Molar mass is typically expressed in units of grams per mole (g/mol). For oxygen, the molar mass is 32.0 g/mol. This is because oxygen naturally occurs as a diatomic molecule, O extsubscript{2}, with each oxygen atom having an atomic mass of approximately 16.0 amu.

So, when using the ideal gas law, the molar mass allows us to relate the number of moles to the actual mass of the gas. It acts as a conversion factor to determine the total mass contained in a certain number of moles, helping us calculate how much of the substance is present in given conditions. When solving problems, always ensure that mass and mole relationships are correctly established using the molar mass for precise results.

Gauge Pressure

Gauge pressure is an important concept when dealing with gases, as it measures the pressure of a gas relative to the atmospheric pressure. This is different from absolute pressure, which includes atmospheric pressure in its measurement.
Here's how it works:

• Gauge pressure is positive when it is greater than atmospheric pressure.
• It is zero when it equals atmospheric pressure.
• When less than atmospheric pressure, the gauge pressure is negative.

In the given problem, the initial gauge pressure of the oxygen in the tank is 3.00 × 10 extsuperscript{5} Pa. To find the true pressure of the gas (known as absolute pressure), one must add the atmospheric pressure (around 1.00 × 10 extsuperscript{5} Pa) to the gauge pressure. Thus, the absolute initial pressure becomes 4.00 × 10 extsuperscript{5} Pa. Understanding gauge pressure is essential for accurately applying the ideal gas law as it directly affects the calculated values of moles and mass of a gas.

Temperature Conversion

Temperature conversion is critical when working with gas laws. Many calculations, especially those involving the ideal gas law, require temperature to be in Kelvin rather than Celsius.

Kelvin is used because it is an absolute temperature scale, starting at absolute zero. In contrast, the Celsius scale doesn't have such a clear physical meaning in terms of kinetic energy of particles, which is what temperature fundamentally measures.

To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature. For example:

• The initial temperature in the problem is 37.0°C. Converted to Kelvin: 37.0 + 273.15 = 310.15 K.
• The final temperature is 22.0°C. Converted to Kelvin: 22.0 + 273.15 = 295.15 K.

These conversions ensure that calculations using the ideal gas law are consistent and accurate.

Gas Leaks

Gas leaks are not only important for safety but also affect calculations regarding the mass and pressure of a gas in a containment system over time.
In the context of the problem, a leak results in a change in the pressure inside an oxygen tank. Initially, we see a high gauge pressure that drops as some gas escapes due to a leak. This naturally leads to a decrease in the number of moles as well as the total mass of the gas inside.
When dealing with gas leaks:

• It's crucial to monitor changes in pressure, which can inform you about how much gas has escaped.
• Comparing initial and final conditions can help determine the amount of gas that has leaked. In the exercise, we observe a reduction from 3.00 × 10 extsuperscript{5} Pa to 1.80 × 10 extsuperscript{5} Pa in gauge pressure.

This allows one to calculate the amount of gas that is still contained versus what has been lost due to the leak, providing a comprehensive understanding of the situation.