What Pressure, In Atm, Would Be Exerted By 0.023 Grams Of Oxygen (O2) If It Occupies 31.6 ML At 91°C?
Understanding the relationship between the amount of gas, its volume, temperature, and pressure is fundamental in chemistry and physics. In this article, we will explore how to determine the pressure exerted by a specific amount of oxygen gas under given conditions. Specifically, we will analyze how 0.023 grams of oxygen (O₂), occupying a volume of 31.6 milliliters at a temperature of 91°C, exerts pressure in atmospheres (atm). Through systematic calculations, we will demonstrate how to approach such problems using the ideal gas law and related concepts.
Understanding the Problem
Before diving into calculations, let’s clarify the problem:
- Mass of oxygen (O₂): 0.023 grams
- Volume occupied: 31.6 mL
- Temperature: 91°C
- Goal: Find the pressure exerted by this oxygen in atm
This problem involves several key concepts:
- Converting mass to moles
- Converting volume to liters
- Converting temperature to Kelvin
- Applying the ideal gas law: PV = nRT
Fundamental Concepts and Constants
To proceed, we need to recall some important constants and conversion factors:
- Molar mass of oxygen (O₂): 32.00 g/mol
- Ideal gas constant (R): 0.082057 atm·L/(mol·K)
- Conversion factors:
- 1 mL = 0.001 L
- Temperature from Celsius to Kelvin: T(K) = T(°C) + 273.15
Step-by-Step Calculation
Let's break down the calculation into manageable steps:
1. Convert mass of oxygen to moles
Number of moles (n) is calculated as:
n = mass / molar mass
n = 0.023 g / 32.00 g/mol
n ≈ 0.00071875 mol
2. Convert volume to liters
Volume in liters (V):
V = 31.6 mL × 0.001 L/mL = 0.0316 L
3. Convert temperature to Kelvin
T = 91°C + 273.15 = 364.15 K
4. Apply the ideal gas law to find pressure
The ideal gas law:
PV = nRT
Rearranged for pressure (P):
P = nRT / V
Substitute the known values:
P = (0.00071875 mol) × (0.082057 atm·L/(mol·K)) × (364.15 K) / 0.0316 L
Calculate numerator:
0.00071875 × 0.082057 × 364.15 ≈ 0.02152
Calculate pressure:
P ≈ 0.02152 / 0.0316 ≈ 0.680 atm
Therefore, the oxygen exerts approximately 0.68 atm of pressure under the given conditions.
Understanding the Significance of the Calculation
Determining the pressure exerted by a specific amount of gas under certain conditions is crucial in various scientific and industrial applications, including:
- Gas storage and transport
- Chemical reactions involving gases
- Respiratory physiology
- Environmental monitoring
This calculation showcases how the ideal gas law provides a straightforward way to relate measurable quantities like mass, volume, and temperature to the pressure exerted by a gas.
Factors Affecting Gas Pressure
While the ideal gas law offers a simplified model, real gases may deviate due to factors like:
- Intermolecular forces
- Gas particle volume
In most laboratory conditions, these deviations are minimal, and the ideal gas approximation remains valid. However, for high pressures or low temperatures, corrections may be necessary.
Additional Considerations and Practical Applications
Practical Applications of Gas Pressure Calculations
- Designing gas storage tanks: Ensuring tanks can withstand the pressure exerted by stored gases.
- Medical applications: Calculating oxygen pressure in medical devices.
- Environmental science: Estimating atmospheric oxygen levels under different conditions.
Tips for Accurate Calculations
- Always convert all quantities to consistent units.
- Use the correct molar mass for the specific gas.
- Convert temperature to Kelvin to avoid errors.
- Double-check calculations, especially when dealing with small or large values.
Summary of the Calculation Process
| Step | Description | Formula / Conversion | Result |
|---|---|---|---|
| 1 | Convert mass to moles | n = 0.023 g / 32 g/mol | ~0.000719 mol |
| 2 | Convert volume to liters | 31.6 mL × 0.001 | 0.0316 L |
| 3 | Convert Celsius to Kelvin | 91 + 273.15 | 364.15 K |
| 4 | Apply ideal gas law | P = nRT / V | ~0.68 atm |
This structured approach ensures clarity and accuracy in solving gas law problems.
Conclusion
By carefully converting all quantities to appropriate units and applying the ideal gas law, we find that 0.023 grams of oxygen occupying 31.6 mL at 91°C exerts approximately 0.68 atm of pressure. This calculation exemplifies fundamental principles of gas behavior and provides insight into how small quantities of gases can exert measurable pressures under specific conditions. Whether in laboratory settings or industrial applications, mastering these calculations is essential for scientists, engineers, and students alike.
Further Reading and Resources
- "Chemistry: The Central Science" by Brown, LeMay, Bursten
- Online calculators for gas law problems
- Educational videos explaining ideal gas law applications
- Research papers on real gas deviations at high pressures
FAQs
Q1: Can the ideal gas law be used for gases at very high pressures?
At very high pressures, gases may deviate from ideal behavior. In such cases, equations of state like the Van der Waals equation provide more accurate results.
Q2: Why is it important to convert temperature to Kelvin?
Because the ideal gas law is based on absolute temperature scales; Kelvin ensures proportionality and correct calculations.
Q3: How accurate is this calculation for real-world scenarios?
For small gases at moderate conditions, the ideal gas law provides a good approximation. Deviations increase at extreme conditions.
Q4: What is the significance of pressure in physical and chemical processes?
Pressure influences reaction rates, phase changes, and the behavior of gases in various systems.
Q5: How can this calculation be adapted for other gases?
Replace the molar mass and use the specific gas constant if necessary, following the same calculation steps.
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By understanding and applying these principles, scientists and students can accurately determine gas pressures under various conditions, enhancing their comprehension of physical chemistry and its applications.
