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Imagine a manometer where air pressure is $\pu{0.8 atm}$ and the gas pressure is more. We move it to sea level which the air pressure is $\pu{1 atm}$. How will the gas pressure change? (the temperature is constant)

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  • $\begingroup$ I'm confused about this gas pressure. If the air pressure is 0.8 atm, how is the gas pressure more? $\endgroup$ – LDC3 Apr 16 '14 at 13:28
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    $\begingroup$ You have a rubber balloon partially inflated to a given diameter... A manometer's readout is a one-dimensional balloon. $\endgroup$ – Uncle Al Apr 16 '14 at 14:23
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Assuming an ideal gas with T constant, $$ PV = nRT $$ the entire right-hand term is constant. Therefore$$ P \propto \frac{1}{V} $$ which is Boyle's law. Since the atmospheric pressure goes up, the liquid portion of the manometer will go down on the side open to the atmosphere, and it will go up on the side open to the gas. Since the volume of the gas goes down, its pressure must increase by Boyle's law.

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