Новое решение
en/5.5.12.md
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| + | ### Statement | ||
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| + | $5.5.12.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | Everty time the pump is connected to the container and the piston is pulled, the volume of the gas increases from $V_0$ to $V_0+V$, and its pressure becomes $V_0/(V_0+V)$ of its former value (assuming constant temperature). In order for the pressure of the gass to drop from $P_0$ to $P$, we must apply $n$ strokes of the pump so that | ||
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| + | \[\left(\frac{V_0}{V_0+V}\right)^n=\frac{P}{P_0}\quad\Rightarrow\quad n=\frac{\ln(P_0/P)}{\ln(1+V/V_0)}.\] | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $5.5.12.$ [Insert the problem statement] | |||
| ### Solution | |||
| Everty time the pump is connected to the container and the piston is pulled, the volume of the gas increases from $V_0$ to $V_0+V$, and its pressure becomes $V_0/(V_0+V)$ of its former value (assuming constant temperature). In order for the pressure of the gass to drop from $P_0$ to $P$, we must apply $n$ strokes of the pump so that | |||
| \[\left(\frac{V_0}{V_0+V}\right)^n=\frac{P}{P_0}\quad\Rightarrow\quad n=\frac{\ln(P_0/P)}{\ln(1+V/V_0)}.\] | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||