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Chapter 5: Problem 7

Would it be easier to drink water with a straw on top of Mt. Everest or at the foot? Explain.

Short Answer

Expert verified
It would be more difficult to drink water with a straw on top of Mount Everest due to the lower atmospheric pressure at that height. This lessened pressure means it would be harder to generate the vacuum necessary to elevate the water within the straw.

Step by step solution

01

Understand the mechanics of drinking with a straw

We start by understanding what happens when water is drawn up through a straw. This is typically achieved by creating a partial vacuum inside the straw, reducing the pressure within the straw. The atmospheric pressure surrounding the straw then pushes the liquid up into the low-pressure region in the straw, thus making the drink rise and allows us to drink it.
02

Consider pressure differences at different heights

Next, we need to think about how pressure differs at different heights. The atmospheric pressure at sea level, or the foot of a mountain, is higher than at the top of a mountain such as Mount Everest. This is because there is a thicker blanket of air pushing down at lower altitudes.
03

Apply the concept to the case of drinking with a straw

Finally, it is crucial to acknowledge that drinking water with a straw becomes harder at higher altitudes. This is because there is lower atmospheric pressure at higher altitudes to push the water up into the straw. Generating the necessary vacuum in the straw would mean reducing an already lower pressure, which is more challenging.

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Most popular questions from this chapter

A sample of air contains only nitrogen and oxygen gases whose partial pressures are 0.80 atm and 0.20 atm, respectively. Calculate the total pressure and the mole fractions of the gases.

A mixture of helium and neon gases is collected over water at \(28.0^{\circ} \mathrm{C}\) and \(745 \mathrm{mmHg}\). If the partial pressure of helium is \(368 \mathrm{mmHg}\), what is the partial pressure of neon? (Vapor pressure of water at \(28^{\circ} \mathrm{C}=\) \(28.3 \mathrm{mmHg} .)\)

Compare the root-mean-square speeds of \(\mathrm{O}_{2}\) and \(\mathrm{UF}_{6}\) at \(65^{\circ} \mathrm{C}\) .

One way to gain a physical understanding of \(b\) in the van der Waals equation is to calculate the "excluded volume." Assume that the distance of closest approach between two similar atoms is the sum of their radii \((2 r) .\) (a) Calculate the volume around each atom into which the center of another atom cannot penetrate. (b) From your result in (a), calculate the excluded volume for 1 mole of the atoms, which is the constant \(b\). How does this volume compare with the sum of the volumes of 1 mole of the atoms?

Convert \(562 \mathrm{mmHg}\) to atm.
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