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Ideal gas law

The ideal gas law states that

p V = N k B T ,

where p is the absolute pressure of a gas, V is the volume it occupies, N is the number of molecules in the gas, and T is its absolute temperature.

The constant k B is called the Boltzmann constant    in honor of the Austrian physicist Ludwig Boltzmann (1844–1906) and has the value

k B = 1.38 × 10 −23 J/K .

The ideal gas law describes the behavior of any real gas when its density is low enough or its temperature high enough that it is far from liquefaction. This encompasses many practical situations. In the next section, we’ll see why it’s independent of the type of gas.

In many situations, the ideal gas law is applied to a sample of gas with a constant number of molecules; for instance, the gas may be in a sealed container. If N is constant, then solving for N shows that pV / T is constant. We can write that fact in a convenient form:

p 1 V 1 T 1 = p 2 V 2 T 2 ,

where the subscripts 1 and 2 refer to any two states of the gas at different times. Again, the temperature must be expressed in kelvin and the pressure must be absolute pressure, which is the sum of gauge pressure and atmospheric pressure.

Calculating pressure changes due to temperature changes

Suppose your bicycle tire is fully inflated, with an absolute pressure of 7.00 × 10 5 Pa (a gauge pressure of just under 90.0 lb/in . 2 ) at a temperature of 18.0 ° C . What is the pressure after its temperature has risen to 35.0 ° C on a hot day? Assume there are no appreciable leaks or changes in volume.

Strategy

The pressure in the tire is changing only because of changes in temperature. We know the initial pressure p 0 = 7.00 × 10 5 Pa , the initial temperature T 0 = 18.0 ° C , and the final temperature T f = 35.0 ° C . We must find the final pressure p f . Since the number of molecules is constant, we can use the equation

p f V f T f = p 0 V 0 T 0 .

Since the volume is constant, V f and V 0 are the same and they divide out. Therefore,

p f T f = p 0 T 0 .

We can then rearrange this to solve for p f :

p f = p 0 T f T 0 ,

where the temperature must be in kelvin.

Solution

  1. Convert temperatures from degrees Celsius to kelvin
    T 0 = ( 18.0 + 273 ) K = 291 K ,

    T f = ( 35.0 + 273 ) K = 308 K .
  2. Substitute the known values into the equation,
    p f = p 0 T f T 0 = 7.00 × 10 5 Pa ( 308 K 291 K ) = 7.41 × 10 5 Pa .

Significance

The final temperature is about 6 % greater than the original temperature, so the final pressure is about 6 % greater as well. Note that absolute pressure (see Fluid Mechanics ) and absolute temperature (see Temperature and Heat ) must be used in the ideal gas law.

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Calculating the number of molecules in a cubic meter of gas

How many molecules are in a typical object, such as gas in a tire or water in a glass? This calculation can give us an idea of how large N typically is. Let’s calculate the number of molecules in the air that a typical healthy young adult inhales in one breath, with a volume of 500 mL, at standard temperature and pressure (STP) , which is defined as 0 ºC and atmospheric pressure. (Our young adult is apparently outside in winter.)

Strategy

Because pressure, volume, and temperature are all specified, we can use the ideal gas law, p V = N k B T , to find N .

Solution

  1. Identify the knowns.
    T = 0 ° C = 273 K , p = 1.01 × 10 5 Pa , V = 500 mL = 5 × 10 −4 m 3 , k B = 1.38 × 10 −23 J/K
  2. Substitute the known values into the equation and solve for N .
    N = p V k B T = ( 1.01 × 10 5 Pa ) ( 5 × 10 −4 m 3 ) ( 1.38 × 10 −23 J/K ) ( 273 K ) = 1.34 × 10 22 molecules

Significance

N is huge, even in small volumes. For example, 1 cm 3 of a gas at STP contains 2.68 × 10 19 molecules. Once again, note that our result for N is the same for all types of gases, including mixtures.

As we observed in the chapter on fluid mechanics, pascals are N/m 2 , so Pa · m 3 = N · m = J . Thus, our result for N is dimensionless, a pure number that could be obtained by counting (in principle) rather than measuring. As it is the number of molecules, we put “molecules” after the number, keeping in mind that it is an aid to communication rather than a unit.

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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