Question

the volume of a gas is changed from 3.6 l to 15.5 l. what is the initial temperature of the gas if the final temperature is 87°c? t = ?°c assume that the number of moles and the pressure remain constant.

Explanation:

Step1: Convert final temp to Kelvin

$T_2 = 87^\circ\text{C} + 273.15 = 360.15\ \text{K}$

Step2: Apply Charles's Law

$\frac{V_1}{T_1} = \frac{V_2}{T_2} \implies T_1 = T_2 \times \frac{V_1}{V_2}$

Step3: Calculate initial temp in Kelvin

$T_1 = 360.15\ \text{K} \times \frac{3.6\ \text{L}}{15.5\ \text{L}} \approx 84.7\ \text{K}$

Step4: Convert to Celsius

$T_1 = 84.7 - 273.15 \approx -188.5^\circ\text{C}$

Answer:

-188.5 (rounded to one decimal place)