JAMB - Physics (2025 - No. 93)
A gas is cooled at a constant pressure from 57ºC was observed to shrink one-fifth (1\5) of its original volume of 2.00cm\(^3\). Find its new temperature
-150\(^o\)C
-207\(^0\)C
-195\(^0\)C
-200\(^0\)C
Explanation
Using Charles' law:
\(\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad \text{(T in Kelvin, constant P)}\)
Calculate \( T_1 \): \(T_1 = 57 + 273 = 330 \, \text{K}\)
Since \( V_2 = \frac{1}{5} V_1 \) (volume shrinks to one-fifth), we find \( T_2 \):
\(T_2 = T_1 \times \frac{V_2}{V_1} = 330 \times \frac{1}{5} = 66 \, \text{K}\)
In °C, this is: \(T_2 = 66 - 273 = -207 \, °C\)
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