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Question

A gas at 2.2atm and 57∘C occupies 5.5L. If the volume is reduced to 3.3L and temperature increased to 107∘C, what is the final pressure?

Options

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Explanation

Given: P1 = 2.2atm, T1 = 57∘C = 330K, V1 = 5.5L, V2 = 3.3L, T2 = 107∘C = 380K. P1V1T1 = P2V2T2. P2 = P1×V1V2×T2T1 = 2.2×5.53.3×380330. P2 = 2.2×1.6667×1.1515≈4.22atm.