If a barometer were built using water \(\left(d=1.0 \mathrm{g} / \mathrm{cm}^{3}\right)\) instead of mercury \(\left(d=13.6 \mathrm{g} / \mathrm{cm}^{3}\right),\) would the column of water be higher than, lower than, or the same as the column of mercury at 1.00 atm? If the level is different, by what factor? Explain.

Short Answer

Expert verified
At 1.00 atm, the column of water in a water-based barometer would be higher than the column of mercury in a mercury-based barometer. The difference between their column heights is due to the different densities of water and mercury. The height of the water column is approximately 10,336 cm, while the height of the mercury column is approximately 76 cm. Therefore, the water column is higher by a factor of approximately 136.

Step by step solution

01

Understand the concept of hydrostatic pressure

In a fluid at rest, the pressure at any given depth depends on fluid density and the height of the column above that point. The hydrostatic pressure is given by the formula: \[P = hρg\], where \(P\) is the hydrostatic pressure, \(h\) is the height of the fluid column, \(ρ\) is the fluid density, and \(g\) is the acceleration due to gravity.
02

Calculate the height for a mercury barometer at 1.00 atm

Using the hydrostatic pressure formula for mercury, we can find the height of the column by replacing pressure with the atmospheric pressure (1.00 atm), and using mercury's density and gravity's acceleration: \[ h_{Hg} = \frac{P_{atm}}{ρ_{Hg} \times g} \] Given that the density of mercury is 13.6 g/cm³, and 1 atm = 101,325 Pa (Pascals) and g = 9.81 m/s², we can now calculate the height: \[ h_{Hg} = \frac{101,325}{13.6 \times 10^{-3} \times 9.81} \] \[ h_{Hg} \approx 76.0 \mathrm{cm} \]
03

Calculate the height for a water barometer at 1.00 atm

Similarly, we will calculate the height of the water column using the same formula: \[ h_{H_2O} = \frac{P_{atm}}{ρ_{H_2O} \times g} \] Given that the density of water is 1.0 g/cm³, we can now calculate the height in a similar fashion as the mercury column: \[ h_{H_2O} = \frac{101,325}{1.0 \times 10^{-3} \times 9.81} \] \[ h_{H_2O} \approx 10,336.0 \mathrm{cm} \]
04

Compare the heights and find the factor by which they differ

Now that we have the heights of both mercury and water columns, we can compare them: \[ h_{H_2O} \approx 10,336.0 \mathrm{cm} \] (Water column) \[ h_{Hg} \approx 76.0 \mathrm{cm} \] (Mercury column) The water column is higher by a considerable amount. To find the factor by which they differ, divide the height of the water column by the height of the mercury column: \[ factor = \frac{h_{H_2O}}{h_{Hg}} \] \[ factor \approx \frac{10,336.0}{76.0} \] \[ factor \approx 136 \] So, at 1.00 atm, the column of water is higher than the mercury column by a factor of approximately 136. This difference is due to the significantly lower density of water compared to mercury, resulting in a much taller column height needed to balance the atmospheric pressure.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Some very effective rocket fuels are composed of lightweight liquids. The fuel composed of dimethylhydrazine \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2}\right]\) mixed with dinitrogen tetroxide was used to power the Lunar Lander in its missions to the moon. The two components react according to the following equation: $$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2}(l)+2 \mathrm{N}_{2} \mathrm{O}_{4}(l) \longrightarrow 3 \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g)$$ If 150 g dimethylhydrazine reacts with excess dinitrogen tetroxide and the product gases are collected at \(127^{\circ} \mathrm{C}\) in an evacuated \(250-\mathrm{L}\) tank, what is the partial pressure of nitrogen gas produced and what is the total pressure in the tank assuming the reaction has \(100 \%\) yield?

At elevated temperatures, sodium chlorate decomposes to produce sodium chloride and oxygen gas. A \(0.8765\)-\(\mathrm{g}\) sample of impure sodium chlorate was heated until the production of oxygen gas ceased. The oxygen gas collected over water occupied \(57.2 \mathrm{mL}\) at a temperature of \(22^{\circ} \mathrm{C}\) and a pressure of \(734\) torr. Calculate the mass percent of \(\mathrm{NaClO}_{3}\) in the original sample. (At \(22^{\circ} \mathrm{C}\) the vapor pressure of water is \(19.8\) torr.)

At STP, \(1.0 \mathrm{L}\) \(Br\) \(_{2}\) reacts completely with \(3.0 \mathrm{L} \mathrm{F}_{2}\), producing \(2.0 \mathrm{L}\) of a product. What is the formula of the product? (All substances are gases.)

Suppose two 200.0 - \(L\) tanks are to be filled separately with the gases helium and hydrogen. What mass of each gas is needed to produce a pressure of 2.70 atm in its respective tank at \(24^{\circ} \mathrm{C} ?\)

We state that the ideal gas law tends to hold best at low pressures and high temperatures. Show how the van der Waals equation simplifies to the ideal gas law under these conditions.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free