To solve this problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature in Kelvin, provided the pressure remains constant. The formula for Charles's Law is:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
Where:
- \( V_1 \) is the initial volume.
- \( T_1 \) is the initial temperature in Kelvin.
- \( V_2 \) is the final volume.
- \( T_2 \) is the final temperature in Kelvin.
Step 1: Convert the temperatures from Celsius to Kelvin.
The conversion from Celsius to Kelvin is done by adding 273.15:
- Initial temperature \( T_1 = 91 , °C = 91 + 273.15 = 364.15 , K \)
- Final temperature \( T_2 = 0 , °C = 0 + 273.15 = 273.15 , K \)
Step 2: Use Charles's Law to solve for the new volume \( V_2 \)
Given:
- \( V_1 = 93 , mL \)
- \( T_1 = 364.15 , K \)
- \( T_2 = 273.15 , K \)
Now, we substitute these values into Charles's Law:
\[ \frac{93 , mL}{364.15 , K} = \frac{V_2}{273.15 , K} \]
Rearranging to solve for \( V_2 \):
\[ V_2 = \frac{93 , mL \times 273.15 , K}{364.15 , K} \]
Step 3: Calculate \( V_2 \)
\[ V_2 = \frac{93 \times 273.15}{364.15} \]
Calculating the numerator:
\[ 93 \times 273.15 = 25333.95 \]
Now, calculating \( V_2 \):
\[ V_2 = \frac{25333.95}{364.15} \approx 69.5 , mL \]
Thus, the new volume \( V_2 \) is approximately 70 mL when rounded to the nearest whole number.
Final Answer: b) 70 mL