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Question

A gas at 2atm and 47∘C occupies 6L. If the temperature drops to −3∘C and pressure becomes 1.5atm, what is the new volume?

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Explanation

Given: P1 = 2atm, T1 = 47∘C = 320K, V1 = 6L, T2 = −3∘C = 270K, P2 = 1.5atm. P1V1T1 = P2V2T2. V2 = V1×P1P2×T2T1 = 6×21.5×270320 = 6×1.3333×0.84375≈6.75L.

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