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Chapter 10: Problem 97

Torricelli, who invented the barometer, used mercury in its construction because mercury has a very high density, which makes it possible to make a more compact barometer than one based on a less dense fluid. Calculate the density of mercury using the observation that the column of mercury is $760 \mathrm{~mm}\( high when the atmospheric pressure is \)1.01 \times 10^{5} \mathrm{~Pa}$. Assume the tube containing the mercury is a cylinder with a constant cross-sectional area.

Short Answer

Expert verified
The density of mercury is approximately \(13,536\text{ kg/m}^3\).

Step by step solution

01

Write down the given parameters

We are given the following values: - Height of mercury column, h: \(760\text{ mm}\) - Atmospheric pressure, P: \(1.01 \times 10^{5}\text{ Pa}\) - Acceleration due to gravity, g: \(9.8 \text{ m/s}^2\) Let's convert the height of the column to meters: \(h = 760\text{ mm} \times \frac{1 \text{ m}}{1000\text{ mm}} = 0.76\text{ m}\)
02

Express the force exerted by the atmosphere

Using the formula for pressure \(P = \frac{F}{A}\), we can express the force exerted by the atmosphere as: \(F_{atm} = P \times A = (1.01 \times 10^{5}\text{ Pa}) \times A\)
03

Express the weight of the mercury column

First, let's express the volume of the mercury column: \(V = A \times h = A \times 0.76\text{ m}\) Next, express the mass of the mercury column using its density, \(\rho\) (which we're trying to find): \(m = \rho \times V = \rho \times (A \times 0.76\text{ m})\) Now we can express the weight of the mercury column, W: \(W = m \times g = (\rho \times (A \times 0.76\text{ m})) \times 9.8 \text{ m/s}^2\)
04

Equate the force exerted by the atmosphere and the weight of the mercury column

As the mercury column is in equilibrium, the force exerted by the atmosphere is equal to the weight of the column: \(F_{atm} = W\) Substitute the expressions for \(F_{atm}\) and W: \((1.01 \times 10^{5}\text{ Pa}) \times A = (\rho \times (A \times 0.76\text{ m})) \times 9.8 \text{ m/s}^2\)
05

Solve for the density of mercury, \(\rho\)

Notice that the cross-sectional area A is present on both sides of the equation, so we can simplify the equation and solve for \(\rho\): \(\rho = \frac{(1.01 \times 10^{5}\text{ Pa})}{(0.76\text{ m} \times 9.8 \text{ m/s}^2)}\) Now, calculate the value of \(\rho\): \(\rho \approx 13,536 \text{ kg/m}^3\) The density of mercury is approximately \(13,536\text{ kg/m}^3\).

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