Answers to Half Sheet Ideal Gas Laws-3 - IDEAL GAS LAW - Name Use ... - Free Printable

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Looking at the problems on the page:

Problem 1: How many moles of gas occupy 98 L at a pressure of 2.8 atm and a temperature of 292 K?

Using PV = nRT
- P = 2.8 atm
- V = 98 L
- T = 292 K
- R = 0.0821 L·atm/(mol·K)

n = PV/RT = (2.8 × 98)/(0.0821 × 292)
n = 274.4/23.9732
n ≈ 11.45 mol

Problem 2: A balloon with 35 L of helium in it is taken from a temperature of 20°C to -35°C. What will the volume be if the pressure remains constant?

This uses Charles's Law: V₁/T₁ = V₂/T₂
- V₁ = 35 L
- T₁ = 20°C = 293 K
- T₂ = -35°C = 238 K

V₂ = V₁ × T₂/T₁ = 35 × 238/293
V₂ = 35 × 0.8123
V₂ ≈ 28.4 L

Problem 3: If 5.0 moles of O₂ and 3.0 moles of N₂ are placed in a 30.0 L tank at a temperature of 25°C, what will the pressure of the resulting mixture of gases be?

Total moles = 5.0 + 3.0 = 8.0 mol
T = 25°C = 298 K
V = 30.0 L

P = nRT/V = (8.0 × 0.0821 × 298)/30.0
P = 195.716/30.0
P ≈ 6.52 atm

Problem 4: A balloon is filled with 35.0 L of helium in the morning when the temperature is 20.0°C. By noon the temperature has risen to 45.0°C. What is the new volume of the balloon?

Charles's Law: V₁/T₁ = V₂/T₂
- V₁ = 35.0 L
- T₁ = 20.0°C = 293 K
- T₂ = 45.0°C = 318 K

V₂ = V₁ × T₂/T₁ = 35.0 × 318/293
V₂ = 35.0 × 1.0853
V₂ ≈ 38.0 L

Problem 5: A balloon that can hold 85 L of air is inflated with 3.5 moles of gas at a pressure of 1.0 atm. What is the temperature in °C of the balloon?

PV = nRT
T = PV/nR = (1.0 × 85)/(3.5 × 0.0821)
T = 85/0.28735
T ≈ 295.8 K

Convert to Celsius: 295.8 - 273 = 22.8°C

Problem 6: CaCO decomposes at 1200°C to form CO₂ gas and CaO. If 25 L of CO₂ are collected at 1200°C, what will the volume of the gas be after it cools to 25°C?

Charles's Law: V₁/T₁ = V₂/T₂
- V₁ = 25 L
- T₁ = 1200°C = 1473 K
- T₂ = 25°C = 298 K

V₂ = V₁ × T₂/T₁ = 25 × 298/1473
V₂ = 25 × 0.2023
V₂ ≈ 5.1 L

Problem 7: We have a 35 L tank that holds helium gas at 120 atm and 22°C. What is the mass of the helium in the tank?

First find moles: PV = nRT
n = PV/RT = (120 × 35)/(0.0821 × 295)
n = 4200/24.2195
n ≈ 173.4 mol

Mass = moles × molar mass
Molar mass of He = 4.0 g/mol
Mass = 173.4 × 4.0 = 693.6 g

Problem 8: An aerosol can containing 400 mL of compressed gas at 5.2 atm. When all of the gas is sprayed into a large plastic bag, the bag inflates to a volume of 2.14 L. What is the pressure of gas inside the bag?

Boyle's Law: P₁V₁ = P₂V₂
- P₁ = 5.2 atm
- V₁ = 400 mL = 0.4 L
- V₂ = 2.14 L

P₂ = P₁V₁/V₂ = (5.2 × 0.4)/2.14
P₂ = 2.08/2.14
P₂ ≈ 0.97 atm

Final Answer:
1. 11.45 moles
2. 28.4 L
3. 6.52 atm
4. 38.0 L
5. 22.8°C
6. 5.1 L
7. 693.6 g (or approximately 694 g)
8. 0.97 atm