Answer:
See Explanations
Explanation:
The Empirical Gas Laws include ...
Boyles Law => P₁V₁ = P₂V₂
Charles Law => T₁/V₁ = T₂/V₂
Gay-Lussac's Law => P₁/T₁= P₂/T₂
Avogadro's Law => V₁/n₁ = V₂/n₂
Combined Gas Law => P₁V₁/T₁ = P₂V₂/T₂
Ideal Gas Law => PV = nRT; R = 0.08206 L·Atm/mol·K
One only needs to learn the 'Combined Gas Law' in that it embodies all of the gas law problems.
Start with a simple table for P, V & T and enter given data. If a variable is not given, assume it is constant. Set up P₁V₁/n₁T₁ = P₂V₂/n₂T₂ and substitute given data. Solve for unknown.
Your Problems:
1. P₁ = 1 Atm P₂ = 3 Atm
V₁ = 300ml V₂ = ?
T₁ = Constant T₂ = T₁
n₁ = Constant n₂ = n₁
P₁V₁/n₁T₁ = P₂V₂/n₂T₂
=> (1Atm)(300ml)/(Constant)(Constant) = (3Atm)(V₂)/(Constant)(Constant)
=> (1Atm)(300ml) = (3Atm)V₂
=> V₂ = (1Atm)(300ml)/(3Atm) = 100ml
2. P₁ = 1 Atm P₂ = ?
V₁ = Constant V₂ = V₁
T₁ = 273K T₂ = 82K
n₁ = Constant n₂ = n₁
P₁V₁/n₁T₁ = P₂V₂/n₂T₂
=> (1Atm)(Constant)/(273K)(Constant) = (P₂)(Constant)/(82K)(Constant)
=> (1Atm)/(273K) = (P₂)/(82K)
=> P₂ = (1Atm)(82K)/(273K)
=> P₂ = 0.3 Atm (1 sig-fig)
3. P₁ = 101.25kPa P₂ = 506kPa
V₁ = 5 Liters V₂ = ?
T₁ = Constant T₂ = T₁
n₁ = Constant n₂ = n₁
P₁V₁/n₁T₁ = P₂V₂/n₂T₂
=> (101.25kPa)(5L)/(Constant)(Constant) = (506kPa)(V₂)/(Constant)(Constant)
=> (101.25kPa)(5L) = (506kPa)(V₂)
=> V₂ = (101.25kPa)(5L)/(506kPa)
=> V₂ = 1.00 Liter
4. P₁ = Constant P₂ = P₁
V₁ = 3.2 Liters V₂ = ?
T₁ = 273°C + 273 = 546K T₂ = 400°C + 273 =673K*
n₁ = Constant n₂ = n₁
*Note => be sure to convert all temps to Kelvin when working empirical gas law problems. K = °C + 273
P₁V₁/n₁T₁ = P₂V₂/n₂T₂
=> (Constant)(3.2L)/(Constant)(546K) = (Constant)(V₂)/(Constant)(673K)
=> (3.2L)/(546K) = (V₂)/(673K)
=> V₂ = (3.2L)(673K)/(546K)
=> V₂ = 3.944 Liters ≅ 3.9 Liters (2 sig-figs)
