Concept:Charles's Law states that the volume of a gas decreases by \( \frac{1}{273.15} \) of its volume at 0°C for every 1 degree Celsius drop in temperature.
Formula:$$ V_t = V_o \left(1 + \frac{t}{273.15}\right) $$
Solution:- If we theoretically cool the gas down to -273.15°C (commonly rounded to -273°C), we substitute this into the equation:
- \( V_t = V_o \left(1 - \frac{273}{273}\right) \).
- \( V_t = V_o (1 - 1) = V_o(0) = 0 \).
- At -273°C (Absolute Zero), the kinetic energy drops to zero, and the theoretical volume of the gas becomes zero (though in reality, the gas will liquefy long before reaching this point).
Why other options are incorrect:- Option B (273°C): At this high temperature, the volume would double compared to its volume at 0°C.
- Options C & D: These are standard warm temperatures where gases occupy large volumes.
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