To solve the problem, we will use Charles' Law, which states that the volume of a gas is directly proportional to its temperature in Kelvin, provided the pressure remains constant. The formula is:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
Where:
- \( V_1 \) is the initial volume of the gas.
- \( T_1 \) is the initial temperature in Kelvin.
- \( V_2 \) is the final volume of the gas.
- \( T_2 \) is the final temperature in Kelvin.
Given:
- Initial volume (\( V_1 \)) = 2.80 L
- Final volume (\( V_2 \)) = 2.7 L
- Final temperature (\( T_2 \)) = 0.00 °C = 273.15 K
Step 1: Convert \( T_2 \) to Kelvin
0.00 °C = 273.15 K
Step 2: Set up the equation to find \( T_1 \):
\[ \frac{2.80 , \text{L}}{T_1} = \frac{2.7 , \text{L}}{273.15 , \text{K}} \]
Step 3: Solve for \( T_1 \):
- Cross-multiply:
\[ 2.80 , \text{L} \times 273.15 , \text{K} = 2.7 , \text{L} \times T_1 \]
- Calculate \( 2.80 \times 273.15 \):
\[ 2.80 \times 273.15 = 764.82 \]
- Then, rearrange the equation to solve for \( T_1 \):
\[ T_1 = \frac{764.82 , \text{L} \cdot \text{K}}{2.7 , \text{L}} \approx 283.60 , \text{K} \]
Step 4: Convert \( T_1 \) to Celsius:
To convert Kelvin back to Celsius, use:
\[ T_C = T_K - 273.15 \]
So,
\[ T_1 (°C) = 283.60 , \text{K} - 273.15 \approx 10.45 , °C \]
Summary:
- Initial Temperature in Kelvin (\( T_1 \)): 283.60 K
- Initial Temperature in Celsius (\( T_1 \)): 10.45 °C
Now, if you have options for completing the list, you can replace the appropriate values based on what I calculated. Those values would be:
- \( T_1 = 283.60 , \text{K} \)
- \( T_1 = 10.45 , °C \)