Perfect (ideal) gas

All gases can be considered approximately as perfect gases at fairly low pressure and fairly high temperature.

Standard conditions(STP) and normal conditions (NTP)

$STP$ : $0^oC $ and $1; bar$. $NTP$ : $20^oC $ and $1; atm$. Note: There is often confusion and some people define: $ ntp $: $ 0^oC $ and $ 1\;atm $. For the approximate calculations this confusion has little importance!

Reminders

Pressure units $ 1 \; bar $ = $ 10^5 \; Pa $

$ 1 \; atm $ = $ 1.0133 \; bar $

$ 760 \; mmHg = 760 \; torr = 1 \; atm$

Temperature units $ T $ = $ t + 273.15$ ($ T $ absolute temperature in Kelvin $ K $, $ t $ temperature in Celsius degrees $^oC $

Number of moles $ n $ = $ \frac{m}{M} $ ($ m $ mass expressed in $ g $, $ M $ molar mass (see →       periodic table ))

The law of Boyle

$ P \cdot V = \; k_1 $ $ k_1 $ is constant for a perfect gas at a given temperature $P_1\cdot V_1=P_2\cdot V_2$ applies to a transformation of a given gas at a given temperature.

The law of Gay-Lussac

$ \frac{V}{T} = \; k_2 $ $ k_2 $ is constant for a perfect gas at a given pressure $\frac{V_1}{T_1}=\frac{V_2}{T_2}$ applies to a transformation of a given gas at a given pressure.

The Law of Charles

$ \frac{P}{T} = \; k_3 $ $ k_3 $ is constant for a perfect gas of given volume $\frac{P_1}{T_1}=\frac{P_2}{T_2}$ applies to a transformation of a given gas at a given volume.

The Law of Avogadro

$ \frac{V}{n} = \;k_4 $ $ k_4 $ is constant for a given perfect gas at a given pressure and temperature. This means that at a given pressure and temperature the same volumes of any gas contain the same number of moles (and therefore molecules). The constant $k_4$ is $22.4 \; L $ at $ 0^oC $ and $ 1\; atm $. So 1 mole of any perfect gas occupies a volume of $ 22.4 \; L $ at $ 0^oC $ and $ 1\; atm $.

The perfect gas law

$ \frac{P \cdot V}{T} $ = $ \; n \cdot R \cdot T $ $ R $ is the universal constant of perfect gases: $ R $ $ = $ $ 0.082 \frac{l \cdot atm}{mol \cdot K} $ If $ P $ is expressed in atmospheres and $ V $ in liters $ R $ $ = $ $ 8.3 \frac{N \cdot m}{mol \cdot K} $ If $ P $ is expressed in Pascal ($ \frac{N}{m^2} $) and $ V $ in $ m^3 $ $\frac{P_1\cdot V_1}{T_1}$ = $\frac{P_2\cdot V_2}{T_2}$ applies to a transformation of a given gas .