To solve this problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its absolute temperature (in Kelvin) when the pressure is constant. Charles's Law is given by the formula:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
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.
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Convert temperatures from Celsius to Kelvin:
- \( T_1 = 91 , ^\circ C + 273.15 = 364.15 , K \)
- \( T_2 = 0 , ^\circ C + 273.15 = 273.15 , K \)
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Apply Charles's Law:
- \( V_1 = 93 , mL \)
- Using the values in the formula:
\[ \frac{93 , mL}{364.15 , K} = \frac{V_2}{273.15 , K} \]
- Cross-multiply to solve for \( V_2 \):
\[ 93 , mL \cdot 273.15 , K = V_2 \cdot 364.15 , K \]
\[ V_2 = \frac{93 , mL \cdot 273.15 , K}{364.15 , K} \]
- Calculate \( V_2 \):
\[ V_2 = \frac{25410.95 , mL \cdot K}{364.15 , K} \approx 69.8 , mL \]
Thus, the new volume of the gas when the temperature is reduced to \( 0 , ^\circ C \) is approximately 69.8 mL.