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Chapter 8: Problem 6

If you put a drinking straw in water, place your finger over the opening, and lift the straw out of the water, some water stays in the straw. Explain.

Short Answer

Expert verified
In short, when you lift a straw with your finger covering the opening, water stays in the straw due to the difference in air pressure inside and outside the straw. The trapped air inside the straw, as per Boyle's law, causes the pressure inside the straw to decrease as you lift it. This lower pressure inside the straw, compared to the constant atmospheric pressure outside, keeps the water from falling out. The finger plays a crucial role in trapping the air and creating this pressure difference.

Step by step solution

01

Identifying forces on the water inside the straw

First, we need to understand the forces acting on the water inside the straw when it's submerged in a liquid and when it's lifted with a finger covering the top. When the straw is submerged in water, there is atmospheric pressure acting on the water inside the straw as well as the water outside the straw. However, once we cover the top of the straw and lift it up, the air pressure inside the straw changes.
02

Understanding air pressure inside the straw

When the top of the straw is covered with a finger and the straw is lifted, the air pressure inside the straw decreases. Since the finger blocks the air from flowing in and out, the volume of air in the straw becomes trapped and is forced to expand as the straw is lifted out of the water. According to Boyle's law, as the volume of a gas increases, its pressure decreases, keeping the product of the pressure and volume constant (P1V1 = P2V2).
03

Comparing air pressure inside and outside the straw

As we lift the straw, the pressure inside the straw has decreased (due to Boyle's law explained in Step 2), while the atmospheric pressure outside the straw remains the same. Since the atmospheric pressure outside the straw is greater than the pressure inside the straw, it pushes the water up the straw and prevents it from falling out.
04

Role of the finger

The finger covering the top of the straw is essential in this process. It traps the air inside the straw and creates the difference in pressure between the inside and outside of the straw. Without the finger, the pressure would equalize as the straw is lifted, and the water would spill out due to gravity.
05

Conclusion

In conclusion, when you put a drinking straw in water, place your finger over the opening, and lift the straw out of the water, some water stays in the straw due to the difference in air pressure inside and outside the straw. The air pressure inside the straw is lower than the atmospheric pressure outside due to the trapped air and Boyle's law. This difference in pressure prevents the water from falling out of the straw, even though it's lifted out of the water.

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Most popular questions from this chapter

A hot-air balloon is filled with air to a volume of \(4.00 \times\) \(10^{3} \mathrm{m}^{3}\) at \(745\) torr and \(21^{\circ} \mathrm{C}\). The air in the balloon is then heated to \(62^{\circ} \mathrm{C},\) causing the balloon to expand to a volume of \(4.20 \times 10^{3} \mathrm{m}^{3} .\) What is the ratio of the number of moles of air in the heated balloon to the original number of moles of air in the balloon? (Hint: Openings in the balloon allow air to flow in and out. Thus the pressure in the balloon is always the same as that of the atmosphere.)

A tank contains a mixture of \(52.5 \mathrm{g}\) oxygen gas and \(65.1 \mathrm{g}\) carbon dioxide gas at \(27^{\circ} \mathrm{C}\). The total pressure in the tank is \(9.21\) atm. Calculate the partial pressures of each gas in the container.

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The oxides of Group \(2 \mathrm{A}\) metals (symbolized by M here) react with carbon dioxide according to the following reaction: $$\mathrm{MO}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{MCO}_{3}(s)$$ A \(2.85\) \(-\mathrm{g}\) sample containing only \(\mathrm{MgO}\) and \(\mathrm{CuO}\) is placed in a 3.00 -L container. The container is filled with \(\mathrm{CO}_{2}\) to a pressure of \(740 .\) torr at \(20 .^{\circ} \mathrm{C}\). After the reaction has gone to completion, the pressure inside the flask is \(390 .\) torr at \(20 .^{\circ} \mathrm{C}\). What is the mass percent of \(MgO\) in the mixture? Assume that only the \(MgO\) reacts with \(\mathrm{CO}_{2}\)

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of \(200 .\) L/min at \(1.50\) atm and ambient temperature. Air is added to the chamber at 1.00 atm and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g),\) three times as much oxygen as is necessary is reacted. Assuming air is \(21\) mole percent \(\mathrm{O}_{2}\) and \(79\) mole percent \(\mathrm{N}_{2}\), calculate the flow rate of air necessary to deliver the required amount of oxygen. b. Under the conditions in part a, combustion of methane was not complete as a mixture of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) was produced. It was determined that \(95.0 \%\) of the carbon in the exhaust gas was present in \(\mathrm{CO}_{2}\). The remainder was present as carbon in \(\mathrm{CO}\). Calculate the composition of the exhaust gas in terms of mole fraction of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Assume \(\mathrm{CH}_{4}\) is completely reacted and \(\mathrm{N}_{2}\) is unreacted.
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