To go from P1V1 = P2V2 to P1/V2 = P2/V1 using middle school algebra, simply:
1. Start with P1V1 = P2V2.
2. Then divide both sides of the equation by V2. (V2 is not zero, of course.) This yields P1V1/V2 = P2.
3. Then divide both sides of this equation by V1. (V1 is not zero, of course.) This yields P1/V2 = P2/V1. Done.
So, the two equations P1V1 = P2V2 and P1/V2 = P2/V1 are equivalent. Use whichever equivalent equation is more convenient at the given moment. (Your choice.)
It's impossible to get from P1V1 = P2V2 to P1/V1 = P2/V2 (except in special, trivial cases). So, P1/V1 = P2/V2 can NOT be an alternate expression of Boyle's Law!
EDIT: The equation P1/V1 = P2/V2 *is* useful, though, in non-Boyle's Law situations. For example, if you want to know how much free gas an Al 80 contains when its SPG reads 500 psig, you can solve the equation
(3000 psig) / (77.4 cu ft) = (500 psig) / V2 for V2. In this case, V2 = 12.9 cu ft.
Safe Diving,
rx7diver
1. Start with P1V1 = P2V2.
2. Then divide both sides of the equation by V2. (V2 is not zero, of course.) This yields P1V1/V2 = P2.
3. Then divide both sides of this equation by V1. (V1 is not zero, of course.) This yields P1/V2 = P2/V1. Done.
So, the two equations P1V1 = P2V2 and P1/V2 = P2/V1 are equivalent. Use whichever equivalent equation is more convenient at the given moment. (Your choice.)
It's impossible to get from P1V1 = P2V2 to P1/V1 = P2/V2 (except in special, trivial cases). So, P1/V1 = P2/V2 can NOT be an alternate expression of Boyle's Law!
EDIT: The equation P1/V1 = P2/V2 *is* useful, though, in non-Boyle's Law situations. For example, if you want to know how much free gas an Al 80 contains when its SPG reads 500 psig, you can solve the equation
(3000 psig) / (77.4 cu ft) = (500 psig) / V2 for V2. In this case, V2 = 12.9 cu ft.
Safe Diving,
rx7diver
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