Ideal Gas Problems: Hydrogen Gas In A Container

Ideal Gas Problems1 A Given Amount Of Hydrogen Gas Is In A Closed E

Analyze various problems related to ideal gases including calculations involving changes in pressure, volume, and temperature; conversions between different units of pressure; and determining amounts of gases in terms of moles, mass, and molecular weight. The problems also include application of the ideal gas law (PV=nRT) to solve for unknowns under given conditions, as well as understanding how gases respond to changes in environmental variables.

Paper For Above instruction

The principles of ideal gas behavior serve as fundamental tools in chemistry for predicting how gases respond to variations in pressure, volume, temperature, and amount. This paper addresses a series of problems that exemplify the application of the ideal gas law, unit conversions, and stoichiometry related to gases. Each problem demonstrates a specific concept, reinforcing the understanding of gas law relationships and their practical implications.

1. Volume Change to Maintain Constant Pressure When Temperature Triples

Considering a fixed amount of hydrogen gas in a closed, expandable container, the first problem explores the relationship between volume and temperature at constant pressure. According to Charles's law, which states that for a fixed amount of gas at constant pressure, volume is directly proportional to temperature (V ∝ T), if temperature triples, volume must also triple to maintain the same pressure. Mathematically, this is expressed as:

V₂ = V₁ (T₂ / T₁). Since T₂ = 3T₁, V₂ = V₁ 3.

2. Effect of Increasing Volume and Doubling Temperature on Pressure

Using Boyle’s law and Gay-Lussac’s law, the second scenario involves increasing the volume to twice its original value and doubling the temperature. The relationship between pressure, volume, and temperature is captured by the combined gas law:

P₁V₁/T₁ = P₂V₂/T₂.

Given V₂ = 2V₁ and T₂ = 2T₁, substituting these into the equation yields:

P₂ = P₁ (V₁ / V₂) (T₂ / T₁) = P₁ (1 / 2) 2 = P₁.

Thus, the pressure remains unchanged under these conditions.

3. Effects of Halving Volume and Tripling Temperature on Pressure

Applying the combined gas law again, with V₂ = V₁ / 2 and T₂ = 3T₁, the new pressure P₂ is:

P₂ = P₁ (V₁ / V₂) (T₂ / T₁) = P₁ 2 3 = 6P₁.

This indicates that the pressure increases sixfold when the volume halves and temperature triples.

4. Conversion of Pressure Units

Conversion between units of pressure involves using specific conversion factors. For example, to convert 4.93 atm to torr, use the conversion factor: 1 atm = 760 torr; thus,

Pressure in torr = 4.93 atm * 760 torr/atm ≈ 3744.8 torr.

Conversely, converting 957 torr to atm involves dividing by 760:

Pressure in atm = 957 torr / 760 ≈ 1.26 atm.

5. Calculating Final Pressure After Compression

Given an initial volume and pressure, and assuming constant temperature, Boyle’s law applies:

P₁V₁ = P₂V₂.

Given V₁=1.00 L, P₁=760 torr, V₂=0.800 L, solving for P₂:

P₂ = P₁ V₁ / V₂ = 760 torr 1.00 / 0.800 = 950 torr.

6. Volume of Gas at Different Conditions

Again, using Boyle’s law, for initial volume V₁=2.00 L at 760 torr, and final volume V₂=?, pressure at 1.25 atm (which is 950 torr), and temperature constant:

V₂ = V₁ P₁ / P₂ = 2.00 L 760 torr / 950 torr ≈ 1.6 L.

7. Determining Temperature Change for Gas Expansion

Using ideal gas law with P, V, and T related as:

V₁/T₁ = V₂/T₂,

where V₁=2.00 L at T₁=27°C=300K, and V₂=3.00 L. The initial pressure is 1.00 atm, and final pressure is 800 torr (~1.053 atm). However, since the pressure change occurs, the more precise approach is to use the combined gas law:

P₁V₁/T₁ = P₂V₂/T₂.

8. Volume Change at Different Conditions

Changing the temperature from 30°C to 0°C (273K), and pressure from 1.00 atm to 800 torr (~1.053 atm), with initial volume 4.00 L, the combined gas law gives:

V₂ = V₁ P₁ / P₂ T₂ / T₁ = 4.00 L 1.00 atm / 1.053 atm 273K / 303K ≈ 3.66 L.

9. Calculating Temperature at a Different Volume and Pressure

Initial data: 2.22 L, 1.00 atm, 60°C (333K). Final conditions: 4.00 L, 1.11 atm, unknown T:

Using P₁V₁/T₁ = P₂V₂/T₂, solving for T₂:

T₂ = P₂V₂ T₁ / P₁V₁ ≈ 1.11 atm 4.00 L 333K / (1.00 atm 2.22 L) ≈ 666 K (approximately 393°C).

10. Volume at STP for Sulfur Dioxide

Given initial volume 4.35 L at 18°C (291K) and 1500 torr (~2 atm), volume at STP (273K, 1 atm):

V₂ = V₁ P₁ / P₂ T₂ / T₁ ≈ 4.35 L 2 atm / 1 atm 273K / 291K ≈ 2.56 L.

11. Gas Volume at STP

Initial conditions: 350 mL at 785 torr (~1.03 atm), find volume at STP:

V₂ = V₁ P₁ / P₂ T₂ / T₁ ≈ 350 mL 1.03 atm / 1 atm 273K / 290K ≈ 330 mL.

12. Moles of N₂O₅ in a Cylinder

Using PV = nRT, with P=1.62 atm, V=5.24 L, T=17°C=290K, R=0.0821 L·atm/(mol·K):

n = PV / RT ≈ (1.62 atm 5.24 L) / (0.0821 290) ≈ 0.358 mol.

13. Mass of Ammonia in a Cylinder

Given P=4.76 atm, V=6.64 L, T=25°C=298K, molar mass of ammonia (NH₃)=17.03 g/mol:

n = PV / RT ≈ (4.76 6.64) / (0.0821 298) ≈ 1.03 mol

Mass = n molar mass ≈ 1.03 mol 17.03 g/mol ≈ 17.56 g.

14. Molecular Mass of Unknown Gas

From given data: mass=0.5396 g, volume=4.000 L, at STP (22.4 L/mol):

Number of moles n = V / 22.4 L/mol ≈ 4.000 / 22.4 ≈ 0.1786 mol.

Thus, molecular mass = mass / n ≈ 0.5396 / 0.1786 ≈ 3.02 g/mol, indicating an extremely light molecule, possibly hydrogen or a similar element or compound.

15. Molecular Mass from Different Data

Mass=0.9243 g, volume=5.000 L at STP:

n = V / 22.4 L/mol ≈ 5.000 / 22.4 ≈ 0.2232 mol, and molecular mass = 0.9243 / 0.2232 ≈ 4.14 g/mol.

This also suggests a light molecule, perhaps involving hydrogen, such as H₂ or related species.

Conclusion

Understanding and applying the ideal gas law and related principles allows chemistry professionals to predict gas behaviors under various conditions with precision. Accurately converting units and solving for different parameters enhances the efficiency of experimental design and data interpretation, which are critical in fields like chemical engineering, environmental science, and physical chemistry.

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