To determine the initial temperature of the gas, we'll use Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure is held constant. The relationship is given by:
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
where:
- [tex]\( V_1 \)[/tex] is the initial volume,
- [tex]\( T_1 \)[/tex] is the initial temperature in Kelvin,
- [tex]\( V_2 \)[/tex] is the final volume, and
- [tex]\( T_2 \)[/tex] is the final temperature in Kelvin.
Given:
- [tex]\( V_1 = 1.21 \)[/tex] L,
- [tex]\( V_2 = 1.0 \)[/tex] L, and
- [tex]\( T_2 = 90^\circ \text{C} \)[/tex].
First, we need to convert the final temperature from Celsius to Kelvin using the formula:
[tex]\[ T(K) = T(C) + 273.15 \][/tex]
So,
[tex]\[ T_2 = 90 + 273.15 = 363.15 \, \text{K} \][/tex]
Using Charles's Law, we can solve for [tex]\( T_1 \)[/tex] in Kelvin:
[tex]\[ T_1 = \frac{V_1 \cdot T_2}{V_2} \][/tex]
Substituting the values:
[tex]\[ T_1 = \frac{1.21 \, \text{L} \cdot 363.15 \, \text{K}}{1.0 \, \text{L}} \][/tex]
[tex]\[ T_1 = 439.4115 \, \text{K} \][/tex]
To convert the initial temperature back to Celsius, we use the formula:
[tex]\[ T(C) = T(K) - 273.15 \][/tex]
So,
[tex]\[ T_1 (C) = 439.4115 - 273.15 = 166.2615^\circ \text{C} \][/tex]
Therefore, the initial temperature of the gas was approximately [tex]\( 166.26^\circ \text{C} \)[/tex].
