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In reviewing the answer supplied by troy, The explanation of his increased water pressure on the larger pipe is as follows: The pressure from the tank is based on the height of the tank. A tank on a 25' tower will supply at least 12.5 pounds per square inch. (we don't know the height of the surface of the water.) The 3/4 inch pipe has an area of .44 sq in. ...


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If the temperature of the gas is kept constant during the compression then the bulk modulus of an ideal gas is just equal to the pressure. The definition of the bulk modulus is: $$ K = -V\frac{dP}{dV} \tag{1} $$ For an ideal gas $PV = RT$, so $P = RT/V$. If the temperature is constant this gives: $$ \frac{dP}{dV} = -\frac{RT}{V^2} \tag{2} $$ and ...


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The Naïve and Wrong Approach Use the ideal gas law, $PV=nRT$, can give you a first-order approximation. This fails, though, because water is not a gas. It does not need to observe the ideal gas law. (In fact, many gasses do not follow the ideal gas law.) Other equations of state could yield a more accurate answer. The application of these equations to this ...


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By just adding on another volume (the attic space), you wouldn't really be applying Charles' Law since you've increased the amount of gas. The temperature would go down though, since you're effectively diluting the hot gas with cooler gas from the attic space. Charles's Law is simply that for a fixed mass of gas at a constant pressure, V and T are ...


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Check out the description of Charles' law on wikipedia. Charles' law relates the temperature and volume of a single body of gas at constant pressure. If you're mixing two bodies of air at different temperatures by punching a hole in the wall between them, you're not changing the volume of a body of gas at constant pressure, you're combining two bodies of ...



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