Question

a gas at 110 °c exerts a pressure of 225 torr in a sealed container. the temperature drops to 65 °c. what is the pressure of the gas at the cooler temperature? p = ? torr hint: be careful with the temperature units.

Explanation:

Step1: Convert temperatures to Kelvin

$T_1 = 110^\circ\text{C} + 273.15 = 383.15\ \text{K}$, $T_2 = 65^\circ\text{C} + 273.15 = 338.15\ \text{K}$

Step2: Apply Gay-Lussac's Law

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

Step3: Calculate final pressure

$P_2 = 225\ \text{torr} \times \frac{338.15}{383.15} \approx 199\ \text{torr}$

Answer:

199