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The general gas equation

  1. A closed gas system initially has a volume of 8 L and a temperature of 100 C. The pressure of the gas is unknown. If the temperature of the gas decreases to 50 C, the gas occupies a volume of 5 L. If the pressure of the gas under these conditions is 1.2 atm, what was the initial pressure of the gas?
  2. A balloon is filled with helium gas at 27 C and a pressure of 1.0 atm. As the balloon rises, the volume of the balloon increases by a factor of 1.6 and the temperature decreases to 15 C. What is the final pressure of the gas (assuming none has escaped)?
  3. 25 cm 3 of gas at 1 atm has a temperature of 20 C. When the gas is compressed to 20 cm 3 , the temperature of the gas increases to 28 C. Calculate the final pressure of the gas.

The ideal gas equation

In the early 1800's, Amedeo Avogadro hypothesised that if you have samples of different gases, of the same volume, at a fixed temperature and pressure, then the samples must contain the same number of freely moving particles (i.e. atoms or molecules).

Avogadro's Law

Equal volumes of gases, at the same temperature and pressure, contain the same number of molecules.

You will remember from an earlier section, that we combined different gas law equations to get one that included temperature, volume and pressure. In this equation, pV = kT, the value of k is different for different masses of gas. If we were to measure the amount of gas in moles, then k = nR, where n is the number of moles of gas and R is the universal gas constant. The value of R is 8.3143 J.K - 1 .mol - 1 , or for most calculations, 8.3 J.K - 1 .mol - 1 . So, if we replace k in the general gas equation, we get the following ideal gas equation .

p V = n R T
  1. The value of R is the same for all gases
  2. All quantities in the equation pV = nRT must be in the same units as the value of R. In other words, SI units must be used throughout the equation.

The following table may help you when you convert to SI units.

Conversion table showing different units of measurement for volume, pressure and temperature.
Variable Pressure ( p ) Volume ( V ) moles ( n ) universal gas constant ( R ) temperature ( K )
SI unit Pascals (Pa) m 3 mol J.K - 1 .mol - 1 Kelvin (K)
Other units and conversions 760 mm Hg = 1 atm = 101325 Pa = 101.325 kPa 1 m 3 = 1000000 cm 3 = 1000 dm 3 = 1000 litres K = C + 273

Khan academy video on the ideal gas law - 1

Khan academy video on the ideal gas law - 2

Two moles of oxygen (O 2 ) gas occupy a volume of 25 dm 3 at a temperature of 40 C. Calculate the pressure of the gas under these conditions.

  1. p = ?

    V = 25 dm 3

    n = 2

    T = 40 C

  2. V = 25 1000 = 0 . 025 m 3
    T = 40 + 273 = 313 K
  3. p V = n R T

    Therefore,

    p = n R T V
  4. p = 2 × 8 . 3 × 313 0 . 025 = 207832 P a = 207 . 8 k P a
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Carbon dioxide (CO 2 ) gas is produced as a result of the reaction between calcium carbonate and hydrochloric acid. The gas that is produced is collected in a 20 dm 3 container. The pressure of the gas is 105 kPa at a temperature of 20 C. What mass of carbon dioxide was produced?

  1. p = 105 kPa

    V = 20 dm 3

    T = 20 C

  2. p = 105 × 1000 = 105000 P a
    T = 20 + 273 = 293 K
    V = 20 1000 = 0 . 02 m 3
  3. p V = n R T

    Therefore,

    n = p V R T
  4. n = 105 000 × 0 . 02 8 . 3 × 293 = 0 . 86 m o l e s
  5. n = m M

    Therefore,

    m = n × M

    The molar mass of CO 2 is calculated as follows:

    M = 12 + ( 2 × 16 ) = 44 g . m o l - 1

    Therefore,

    m = 0 . 86 × 44 = 37 . 84 g
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Source:  OpenStax, Siyavula textbooks: grade 11 physical science. OpenStax CNX. Jul 29, 2011 Download for free at http://cnx.org/content/col11241/1.2
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