A given mass of gas occupies 2dm3 at 300k. At what temperature will its volume be doubled, keeping the pressure constant?
The correct answer is D. 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
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