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80 points to whoever ACTUALLY solves ALL 4!!

It has something to do with boyle’s/charle’s/G-L/Combined Gas Laws

I’m super lost and I’ve been working on this for nearly 2 hours!

80 points to whoever ACTUALLY solves ALL 4!! It has something to do with boyle’s/charle-example-1

1 Answer

4 votes

Answer:

See Explanations

Step-by-step 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)

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