Question

a gas in a sealed container at 86.0 °c decreases in temperature to 30.0 °c where it has a pressure of 1.80 atm. what was the initial pressure of the gas?
p = ? atm

Explanation:

Step1: Convert temperatures to Kelvin

Initial temp: $T_1 = 86.0 + 273.15 = 359.15$ K
Final temp: $T_2 = 30.0 + 273.15 = 303.15$ K

Step2: Apply Gay-Lussac's Law

Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ → $P_1 = P_2 \times \frac{T_1}{T_2}$

Step3: Calculate initial pressure

$P_1 = 1.80 \times \frac{359.15}{303.15} ≈ 2.13$ atm

Answer:

2.13