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Chapter 19: Problem 29

An auditorium has dimensions \(10.0 \mathrm{m} \times 20.0 \mathrm{m} \times 30.0 \mathrm{m}\) How many molecules of air fill the auditorium at \(20.0^{\circ} \mathrm{C}\) and a pressure of \(101 \mathrm{kPa} ?\)

Short Answer

Expert verified
The number of air molecules filling the auditorium at the given conditions is calculated using the volume of the auditorium, temperature, pressure, ideal gas law and Avogadro's number.

Step by step solution

01

Calculate the volume of the auditorium in cubic meters

First, find the volume of the auditorium by multiplying its dimensions. The volume V in cubic meters is given by \( V = length \times width \times height = 10.0m \times 20.0m \times 30.0m = 6000.0m^{3} \)
02

Convert temperature into Kelvin

Next, convert the temperature from Celsius to Kelvin. The conversion is \( T(K) = T(掳C) + 273.15 = 20.0掳C + 273.15 = 293.15K \)
03

Convert pressure into Pa

Now, convert the pressure from kPa to Pa by multiplying the pressure value by 1000. So, \( P = 101kPa \times 1000 = 101000Pa \)
04

Calculate the number of moles using Ideal Gas Law

Then, use the ideal gas law to find the number of moles. Rearrange the ideal gas law equation to solve for n: \( n = \frac{PV}{RT}= \frac{101000Pa \times 6000.0m^{3}}{8.3145\frac{m^2 kg}{s^2 K mol} \times 293.15K} \)
05

Calculate number of molecules in auditorium

Finally, multiply the number of moles by Avogadro鈥檚 number \((6.022 \times 10^{23} molecules/mol) \) to find the number of molecules. \( molecules = n \times Avogadro's \ Number \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume calculation
To find out how much air fills a space, we first need to calculate the volume of that space. In the case of an auditorium, which is often a large rectangular prism, we do this by multiplying its length, width, and height.
Consider the dimensions provided: 10.0 meters in height, 20.0 meters in width, and 30.0 meters in length. When you multiply these three dimensions, the total volume of the auditorium is 6000 cubic meters:
  • Length = 10.0 m
  • Width = 20.0 m
  • Height = 30.0 m
  • Volume, \( V = 10.0 \times 20.0 \times 30.0 = 6000.0 m^3 \)
By understanding volume calculation, we find the auditorium has room for 6000 cubic meters of air.
Temperature conversion
Temperature conversion is key when using the Ideal Gas Law, which requires temperatures to be in Kelvin. To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature.
In this problem, the temperature inside the auditorium is given as 20.0掳C. To convert it to Kelvin, add 273.15:
  • Temperature in 掳C = 20.0
  • Temperature in Kelvin = \(20.0 + 273.15 = 293.15 \) K
By converting to Kelvin, we make temperature compatible with the gas law formula, enabling accurate calculations.
Pressure conversion
Pressure must also be in a standard unit, which for the Ideal Gas Law is Pascals (Pa). Pressure given in kilopascals (kPa) must be converted to Pascals by multiplying by 1000.
Here, the auditorium's air pressure is reported as 101 kPa. The conversion to Pascals is straightforward:
  • Pressure in kPa = 101
  • Pressure in Pascals = \(101 \times 1000 = 101000 \) Pa
Converting pressure ensures that all units are consistent, leading to correct results when applying the Ideal Gas Law.
Number of moles
The Ideal Gas Law equation is essential for determining the number of moles inside the auditorium. The equation is often written as: \[ PV = nRT \] Where:
  • \( P \) is the pressure in Pascals
  • \( V \) is the volume in cubic meters
  • \( n \) is the number of moles
  • \( R \) is the ideal gas constant (8.3145 \( \frac{m^2 kg}{s^2 K mol} \))
  • \( T \) is the temperature in Kelvin
To find \( n \), solve the equation for \( n \): \[ n = \frac{PV}{RT} \] Substituting our known values:
  • \( n = \frac{101000 \times 6000.0}{8.3145 \times 293.15} \)
By solving, we find the number of moles in the auditorium. This represents the quantity of gas particles under the specified conditions of volume, temperature, and pressure.
Avogadro's number
Avogadro鈥檚 number connects the number of moles to the number of molecules. It represents the number of constituent particles, usually atoms or molecules, that are contained in one mole. The value is approximately \( 6.022 \times 10^{23} \) particles/mol.

Once you know the number of moles using the Ideal Gas Law, determine the total number of molecules by multiplying the moles by Avogadro鈥檚 number:
  • \( ext{Number of Molecules} = n \times 6.022 \times 10^{23} \)
This calculation gives the total number of molecules of air in the auditorium. It is an immense number, illustrating just how many molecules fill even a seemingly limited space!

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Most popular questions from this chapter

The density of gasoline is \(730 \mathrm{kg} / \mathrm{m}^{3}\) at \(0^{\circ} \mathrm{C} .\) Its average coefficient of volume expansion is \(9.60 \times 10^{-4} /^{\circ} \mathrm{C} .\) If 1.00 gal of gasoline occupies \(0.00380 \mathrm{m}^{3},\) how many extra kilograms of gasoline would you get if you bought \(10.0 \mathrm{gal}\) of gasoline at \(0^{\circ} \mathrm{C}\) rather than at \(20.0^{\circ} \mathrm{C}\) from a pump that is not temperature compensated?

Helium gas is sold in steel tanks. If the helium is used to inflate a balloon, could the balloon lift the spherical tank the helium came in? Justify your answer. Steel will rupture if subjected to tensile stress greater than its yield strength of \(5 \times 10^{8} \mathrm{N} / \mathrm{m}^{2} .\) Suggestion: You may consider a steel shell of radius \(r\) and thickness \(t\) containing helium at high pressure and on the verge of breaking apart into two hemispheres.

The pressure gauge on a tank registers the gauge pressure, which is the difference between the interior and exterior pressure. When the tank is full of oxygen \(\left(\mathrm{O}_{2}\right),\) it contains \(12.0 \mathrm{kg}\) of the gas at a gauge pressure of \(40.0 \mathrm{atm} .\) Determine the mass of oxygen that has been withdrawn from the tank when the pressure reading is 25.0 atm. Assume that the temperature of the tank remains constant.

(a) Use the equation of state for an ideal gas and the definition of the coefficient of volume expansion, in the form \(\beta=(1 / V) d V / d T,\) to show that the coefficient of volume expansion for an ideal gas at constant pressure is given by \(\bar{\beta}=1 / T,\) where \(T\) is the absolute temperature. (b) What value does this expression predict for \(\beta\) at \(0^{\circ} \mathrm{C} ?\) Compare this result with the experimental values for helium and air in Table \(19.1 .\) Note that these are much larger than the coefficients of volume expansion for most liquids and solids.

A diving bell in the shape of a cylinder with a height of \(2.50 \mathrm{m}\) is closed at the upper end and open at the lower end. The bell is lowered from air into sea water \(\left(\rho=1.025 \mathrm{g} / \mathrm{cm}^{3}\right) .\) The air in the bell is initially at \(20.0^{\circ} \mathrm{C} .\) The bell is lowered to a depth (measured to the bottom of the bell) of 45.0 fathoms or \(82.3 \mathrm{m}\). At this depth the water temperature is \(4.0^{\circ} \mathrm{C},\) and the bell is in thermal equilibrium with the water. (a) How high does sea water rise in the bell? (b) To what minimum pressure must the air in the bell be raised to expel the water that entered?
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