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Chapter 13: Problem 37

Air at room temperature and atmospheric pressure has a mass density of $1.2 \mathrm{kg} / \mathrm{m}^{3} .$ The average molecular mass of air is 29.0 u. How many molecules are in \(1.0 \mathrm{cm}^{3}\) of air?

Short Answer

Expert verified
Answer: Approximately \(2.49 \times 10^{19}\) air molecules are present in 1.0 cm鲁 of air.

Step by step solution

01

Convert mass density of air

First, we need to convert the mass density of air from kg/m鲁 to g/cm鲁. We know that 1 kg = 1000 g and 1 m = 100 cm. Therefore, \(1.2 \: \text{kg/m}^3 \times \frac{1000 \: \text{g}}{1 \: \text{kg}} \times \left(\frac{1 \: \text{m}}{100 \: \text{cm}}\right)^3 = 1.2 \times 10^{-3} \: \text{g/cm}^3\)
02

Calculate total mass in 1 cm鲁

Now, let's find the total mass of 1 cm鲁 of air by multiplying the mass density by the volume: \(1.2 \times 10^{-3} \: \text{g/cm}^3 \times 1.0 \: \text{cm}^3 = 1.2 \times 10^{-3} \: \text{g}\)
03

Calculate moles of air molecules

Since the average molecular mass of air is given as 29.0 u, we can use this to calculate the moles of air molecules in 1 cm鲁 of air. To do this, we need to divide the total mass of air molecules by the average molecular mass: \(\text{Moles of air molecules} = \frac{\text{Total mass of air molecules}}{\text{Average molecular mass of air}}\) \(\text{Moles of air molecules} = \frac{1.2 \times 10^{-3} \: \text{g}}{29.0 \: \text{g/mol}} = 4.14 \times 10^{-5} \: \text{mol}\)
04

Use Avogadro's number to find the number of air molecules

Finally, we can use Avogadro's number (\(6.022 \times 10^{23} \: \text{molecules/mol}\)) to convert the moles of air molecules into the actual number of molecules: \(\text{Number of air molecules} = \text{Moles of air molecules} \times \text{Avogadro's number}\) \(\text{Number of air molecules} = 4.14 \times 10^{-5} \: \text{mol} \times 6.022 \times 10^{23} \: \text{molecules/mol} \approx 2.49 \times 10^{19} \: \text{molecules}\) So, there are approximately \(2.49 \times 10^{19}\) air molecules in 1.0 cm鲁 of air.

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Most popular questions from this chapter

A bubble rises from the bottom of a lake of depth \(80.0 \mathrm{m},\) where the temperature is \(4^{\circ} \mathrm{C} .\) The water temperature at the surface is \(18^{\circ} \mathrm{C} .\) If the bubble's initial diameter is \(1.00 \mathrm{mm},\) what is its diameter when it reaches the surface? (Ignore the surface tension of water. Assume the bubble warms as it rises to the same temperature as the water and retains a spherical shape. Assume \(\left.P_{\mathrm{atm}}=1.0 \mathrm{atm} .\right)\)
Agnes Pockels \((1862-1935)\) was able to determine Avogadro's number using only a few household chemicals, in particular oleic acid, whose formula is \(\mathrm{C}_{18} \mathrm{H}_{34} \mathrm{O}_{2}\) (a) What is the molar mass of this acid? (b) The mass of one drop of oleic acid is \(2.3 \times 10^{-5} \mathrm{g}\) and the volume is $2.6 \times 10^{-5} \mathrm{cm}^{3} .$ How many moles of oleic acid are there in one drop? (c) Now all Pockels needed was to find the number of molecules of oleic acid. Luckily, when oleic acid is spread out on water, it lines up in a layer one molecule thick. If the base of the molecule of oleic acid is a square of side \(d\), the height of the molecule is known to be \(7 d .\) Pockels spread out one drop of oleic acid on some water, and measured the area to be \(70.0 \mathrm{cm}^{2}\) Using the volume and the area of oleic acid, what is \(d ?\) (d) If we assume that this film is one molecule thick, how many molecules of oleic acid are there in the drop? (e) What value does this give you for Avogadro's number?
As a Boeing 747 gains altitude, the passenger cabin is pressurized. However, the cabin is not pressurized fully to atmospheric $\left(1.01 \times 10^{5} \mathrm{Pa}\right),$ as it would be at sea level, but rather pressurized to \(7.62 \times 10^{4} \mathrm{Pa}\). Suppose a 747 takes off from sea level when the temperature in the airplane is \(25.0^{\circ} \mathrm{C}\) and the pressure is \(1.01 \times 10^{5} \mathrm{Pa} .\) (a) If the cabin temperature remains at \(25.0^{\circ} \mathrm{C},\) what is the percentage change in the number of moles of air in the cabin? (b) If instead, the number of moles of air in the cabin does not change, what would the temperature be?
Show that the rms speed of a molecule in an ideal gas at absolute temperature \(T\) is given by $$v_{\mathrm{rms}}=\sqrt{\frac{3 k T}{m}}$$ where \(k\) is Boltzmann's constant and \(m\) is the mass of a molecule.
In intergalactic space, there is an average of about one hydrogen atom per \(\mathrm{cm}^{3}\) and the temperature is \(3 \mathrm{K}\) What is the absolute pressure?
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