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You're taking outside air, compressing it to more than double the pressure, ($14.96+15$, compared to atmospheric $14.96$) and putting it into a 9 cubit foot container. So the pump must suck in around 18 cubic feet of outside air and run it into the container. If we assume that the quoted figure is the through-put of the pump with no back pressure, and that ...

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Actually, it's not as easy as that. In 3.6 minutes, you have filled the space, but there is no additional pressure. You need to run the pump as long again to get an atmosphere of pressure inside the paddle board. Also, as the pressure increases in the paddle board, the efficiently of the pump decreases, therefore taking longer to fill it to the correct ...

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I regularly use boiling water as a timer. I set my whistling tea kettle to boil first thing in the morning, then go to the bathroom to brush my teeth and wash my face. I know that if I put 3 cups in the pot, I have approximately six minutes to complete my washing up; if I use 4 cups, I have 8 minutes. So, about 2 minutes per cup of water. (I always measure ...

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Answer to question (1) is NO: Using the equations of the comments $(P_1 V_1=P_2 V_2$), you get: $P_2=P_1/(1+.06666)$ (in this case of a small drop in pressure the answer will be approximately correct though, because $1/(1+x) \simeq 1-x$ for $x<<1$) Answer to question (2) is YES: Using $n_2RT_1=(1-0.06666)n_1RT_1$ you get $P_2=(1-.0666)P_1$

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