A given mass of gas occupies 2dm3 at 300k. At what temperature will its volume be doubled, keeping the pressure constant?
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Correct Answer: Option D
Explanation:
At constant pressure connotes Charle's law
\(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\)
\(\frac{2dm^3}{300k}\) = 2 x \(\frac{2dm^3}{T_2}\)
T2= 600k
At constant pressure connotes Charle's law
\(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\)
\(\frac{2dm^3}{300k}\) = 2 x \(\frac{2dm^3}{T_2}\)
T2= 600k