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Chapter 6: Problem 4

Calculate the height of a column of liquid glycerol \(\overline{\left(d=1.26 \mathrm{g} / \mathrm{cm}^{3}\right), \text { in meters, required to exert the }}\) same pressure as \(3.02 \mathrm{m}\) of \(\mathrm{CCl}_{4}(\mathrm{l})\left(d=1.59 \mathrm{g} / \mathrm{cm}^{3}\right)\)

Short Answer

Expert verified
The height of glycerol column required to exert the same pressure as 3.02 m of CCl4 is approximately 3.81 m.

Step by step solution

01

Convert densities to SI units

The densities of glycerol and CCl4 are given in g/cm^3. To convert them to SI units (kg/m^3), we need to multiply the given densities by 1000. So, the densities becomes: \(d_{glycerol} = 1.26 * 1000 kg/m^3\), \(d_{CCl4} = 1.59 * 1000 kg/m^3\)
02

Convert height of CCl4 to meters

The height of the CCl4 column is given in metres, so no conversion is required.
03

Set up the pressure equation

We will use the formula for pressure in fluids: Pressure = density * gravity * height. Since the pressures exerted by both columns are equal, we can write: \(d_{glycerol} * g * h_{glycerol} = d_{CCl4} * g * h_{CCl4}\) where g is the acceleration due to gravity (9.81 m/s^2), \(h_{glycerol}\) is the height of the glycerol column, and \(h_{CCl4}\) is the height of the CCl4 column (3.02 m). The g factor can be cancelled from both sides.
04

Solve for height of glycerol column

Now, we can rearrange the formula and solve for \(h_{glycerol}\): \(h_{glycerol} = \frac{d_{CCl4} * h_{CCl4}}{d_{glycerol}}\). Substituting the known values: \(h_{glycerol} = \frac{1.59 * 1000 * 3.02}{1.26 * 1000} m\). To calculate the result, we multiply and divide the numbers.

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Most popular questions from this chapter

Calculate \(u_{\mathrm{rms}},\) in meters per second, for \(\mathrm{Cl}_{2}(\mathrm{g})\) molecules at \(30^{\circ} \mathrm{C}\)

If the van der Waals equation is solved for volume, a cubic equation is obtained. (a) Derive the equation below by rearranging equation (6.26). \(V^{3}-n\left(\frac{R T+b P}{P}\right) V^{2}+\left(\frac{n^{2} a}{P}\right) V-\frac{n^{3} a b}{P}=0\) (b) What is the volume, in liters, occupied by \(185 \mathrm{g}\) \(\mathrm{CO}_{2}(\mathrm{g})\) at a pressure of \(125 \mathrm{atm}\) and \(286 \mathrm{K} ?\) For \(\mathrm{CO}_{2}(\mathrm{g})\) \(a=3.61 \mathrm{L}^{2} \mathrm{atm} \mathrm{mol}^{-2}\) and \(b=0.0429 \mathrm{Lmol}^{-1}\) [Hint: Use the ideal gas equation to obtain an estimate of the volume. Then refine your estimate, either by trial and error, or using the method of successive approximations. See Appendix A, pages A5-A6, for a description of the method of successive approximations.

A sample of \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(23^{\circ} \mathrm{C}\) and a barometric pressure of 751 Torr. The vapor pressure of water at \(23^{\circ} \mathrm{C}\) is \(21 \mathrm{mmHg}\). The partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in the sample collected is (a) \(21 \mathrm{mmHg}_{i}\) (b) 751 Torr; \((\mathrm{c}) 0.96 \mathrm{atm} ;\) (d) \(1.02 \mathrm{atm}\).

A gaseous hydrocarbon that is \(82.7 \%\) C and \(17.3 \%\) H by mass has a density of \(2.35 \mathrm{g} / \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) and 752 Torr. What is the molecular formula of this hydrocarbon?

A sample of \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(24^{\circ} \mathrm{C}\) The volume of gas is 1.16 L. In a subsequent experiment, it is determined that the mass of \(\mathrm{O}_{2}\) present is 1.46 g. What must have been the barometric pressure at the time the gas was collected? (Vapor pressure of water \(=22.4 \text { Torr. })\)
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