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: $T_1 = 86.0^\circ\text{C} + 273.15 = 359.15\ \text{K}$
Final temp: $T_2 = 30.0^\circ\text{C} + 273.15 = 303.15\ \text{K}$

Step2: Apply Gay-Lussac's Law

$\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies P_1 = P_2 \cdot \frac{T_1}{T_2}$

Step3: Calculate initial pressure

$P_1 = 1.80\ \text{atm} \cdot \frac{359.15}{303.15} \approx 2.13\ \text{atm}$

Answer:

2.13