Question

1. choose the the answer.

a gas in a closed container is pressurized from 12.0 atm to 15.0 atm. if the final temperature is recorded as 45.0 °c, what was the celsius initial temperature of the gas?
○ 254 °c
○ 56 °c
○ -19 °c

Explanation:

Step1: Convert temp to Kelvin

$T_2 = 45.0 + 273.15 = 318.15\ \text{K}$

Step2: Apply Gay-Lussac's Law

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

Step3: Substitute values to find $T_1$

$T_1 = \frac{12.0\ \text{atm} \times 318.15\ \text{K}}{15.0\ \text{atm}} = 254.52\ \text{K}$

Step4: Convert back to Celsius

$T_{1(\text{C})} = 254.52 - 273.15 = -18.63\ ^\circ\text{C} \approx -19\ ^\circ\text{C}$

Answer:

-19 °C