A sample of argon initially has a volume of 5.0 L and the pressure is 2 atm. If the final temperature is 30° C, the final volume is 6 L, and the final pressure is 8atm, what was the initial temperature of the argon?
(P1V1/T1) = (P2V2/T2)
Post your work if you get stuck.
REMEMBER that T must be in kelvin.
3465.k
To find the initial temperature of the argon, we can use the combined gas law equation. The combined gas law relates the initial and final states of a gas sample under constant amount and constant conditions.
The combined gas law equation is:
(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂
Where:
- P₁ and P₂ are the initial and final pressures, respectively
- V₁ and V₂ are the initial and final volumes, respectively
- T₁ and T₂ are the initial and final temperatures in Kelvin, respectively
Given in the problem:
- P₁ = 2 atm
- V₁ = 5.0 L
- T₂ = 30°C = 30 + 273.15 = 303.15 K
- V₂ = 6 L
- P₂ = 8 atm
Now, let's substitute these values into the equation and solve for T₁:
(2 atm * 5.0 L) / T₁ = (8 atm * 6 L) / 303.15 K
To isolate T₁, we need to cross multiply and rearrange the equation:
(2 atm * 5.0 L * 303.15 K) = (8 atm * 6 L * T₁)
Now, divide both sides by (8 atm * 6 L) to solve for T₁:
T₁ = (2 atm * 5.0 L * 303.15 K) / (8 atm * 6 L)
T₁ simplifies to:
T₁ ≈ 252.63 K
Therefore, the initial temperature of the argon was approximately 252.63 Kelvin.