If pressure decreases from 2 atm to 1 atm at constant temperature, what happens to the volume if the initial volume is V0?

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Multiple Choice

If pressure decreases from 2 atm to 1 atm at constant temperature, what happens to the volume if the initial volume is V0?

Explanation:
At constant temperature, the product of pressure and volume stays the same (Boyle’s law): P1V1 = P2V2. With P1 = 2 atm and V1 = V0, and P2 = 1 atm, solve for V2: V2 = (P1V1)/P2 = (2 × V0)/1 = 2V0. So the volume must double to keep P×V constant when pressure halves. Intuitively, reducing pressure lets the gas occupy more space, so the volume increases until the product P×V returns to its original value. The volume becomes twice the initial, V2 = 2V0.

At constant temperature, the product of pressure and volume stays the same (Boyle’s law): P1V1 = P2V2. With P1 = 2 atm and V1 = V0, and P2 = 1 atm, solve for V2: V2 = (P1V1)/P2 = (2 × V0)/1 = 2V0. So the volume must double to keep P×V constant when pressure halves. Intuitively, reducing pressure lets the gas occupy more space, so the volume increases until the product P×V returns to its original value. The volume becomes twice the initial, V2 = 2V0.

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