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Chapter 22: Problem 31

Estimate the van der Waals constant \(b\) for \(\mathrm{H}_{2} \mathrm{O}\) knowing that one kilogram of water has a volume of \(0.001 \mathrm{~m}^{3}\). The molar mass of water is \(18 \mathrm{~g} / \mathrm{mol}\).

Short Answer

Expert verified
The estimated van der Waals constant \(b\) for water is approximately \(1.8 \times 10^{-5} \mathrm{~m}^{3/mol}\).

Step by step solution

01

Determine the number of moles in 1 kilogram of water

First, convert the mass of water from kilograms to grams (since the molar mass is given in grams). Since \(1 \mathrm{~kg} = 1000 \mathrm{~g}\), there are \(1000 \mathrm{~g}\) of water in one kilogram. The number of moles of water can be found using the molar mass \(M\), by applying the following formula: \[N = \frac{m}{M}\]where:\(m\) = mass of the substance = 1000 g\(N\) = number of moles\(M\) = molar mass of the substance = 18 g/mol
02

Calculate the number of moles

Substitute the given values into the formula:\[N = \frac{1000 \mathrm{~g}}{18 \mathrm{~g/mol}} = 55.56 \mathrm{~mol}\]This means there are approximately 55.56 moles of water in one kilogram.
03

Calculate the van der Waals constant

The volume occupied by one mole of a real gas at zero pressure according to the van der Waals equation is the van der Waals constant \(b\). As the total volume of the water is given, \(b\) can be calculated by dividing the volume by the number of moles:\[b = \frac{V_{total}}{N}\]where:\(V_{total}\) = total volume = 0.001 m^3\(N\) = number of moles = 55.56 mol
04

Final Calculation

Substitute the given values into the formula:\[b = \frac{0.001 \mathrm{~m}^{3}}{55.56 \mathrm{~mol}} = 1.8 \times 10^{-5} \mathrm{~m}^{3/mol}\]So, the estimated van der Waals constant \(b\) for water is approximately \(1.8 \times 10^{-5} \mathrm{~m}^{3/mol}\).

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Most popular questions from this chapter

The mass of the \(\mathrm{H}_{2}\) molecule is \(3.3 \times 10^{-24} \mathrm{~g}\). If \(1.6 \times 10^{23}\) hydrogen molecules per second strike \(2.0 \mathrm{~cm}^{2}\) of wall at an angle of \(55^{\circ}\) with the normal when moving with a speed of \(1.0 \times 10^{5} \mathrm{~cm} / \mathrm{s}\), what pressure do they exert on the wall?

Consider a sample of argon gas at \(35.0^{\circ} \mathrm{C}\) and \(1.22 \mathrm{~atm}\) pressure. Suppose that the radius of a (spherical) argon atom is \(0.710 \times 10^{-10} \mathrm{~m} .\) Calculate the fraction of the container volume actually occupied by atoms.

A steel tank contains \(315 \mathrm{~g}\) of ammonia gas \(\left(\mathrm{NH}_{3}\right)\) at an absolute pressure of \(1.35 \times 10^{6} \mathrm{~Pa}\) and temperature \(77.0^{\circ} \mathrm{C} .(a)\) What is the volume of the tank? \((b)\) The tank is checked later when the temperature has dropped to \(22.0^{\circ} \mathrm{C}\) and the absolute pressure has fallen to \(8.68 \times 10^{5} \mathrm{~Pa}\). How many grams of gas leaked out of the tank?

(a) Ten particles are moving with the following speeds: four at \(200 \mathrm{~m} / \mathrm{s}\), two at \(500 \mathrm{~m} / \mathrm{s}\), and four at \(600 \mathrm{~m} / \mathrm{s}\). Calculate the average and root-mean-square speeds. Is \(v_{\mathrm{rms}}>v_{\mathrm{av}} ?(b)\) Make up your own speed distribution for the ten particles and show that \(v_{\mathrm{rms}} \geq v_{\mathrm{av}}\) for your distribution. ( \(c\) ) Under what condition (if any) does \(v_{\mathrm{rms}}=v_{\mathrm{av}} ?\)

Calculate the root-mean-square speed of ammonia \(\left(\mathrm{NH}_{3}\right)\) molecules at \(56.0^{\circ} \mathrm{C}\). An atom of nitrogen has a mass of \(2.33 \times 10^{-26} \mathrm{~kg}\) and an atom of hydrogen has a mass of \(1.67 \times 10^{-27} \mathrm{~kg}\)
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