Ideal gas law/Tutorials

• All gases mentioned below are assumed to be ideal, i.e. their p, V, T dependence is given by the ideal gas law.

• Absolute temperature is given by K = °C + 273.15.

• All pressures are absolute.

• The molar gas constant R = 0.082057 atm·L/(K·mol).

Example problems

See also the tutorials on the Video subpage.

Problem 1

Determine the volume of 1 mol of ideal gas at pressure 1 atm and temperature 20 °C.

$V={\frac {n\,R\,T}{p}}={\frac {1\cdot 0.082057\cdot (20+273.15)}{1}}\quad \left[{\frac {\mathrm {mol} \cdot {\frac {\mathrm {atm} \cdot \mathrm {L} }{\mathrm {K} \cdot \mathrm {mol} }}\cdot \mathrm {K} }{\mathrm {atm} }}\right]=24.0550\quad [\mathrm {L} ]$

Problem 2

Compute from Charles' and Gay-Lussac's law (V/T is constant) the volume of an ideal gas at 1 atm and 0 °C (Use the final result of the previous problem). Write V_T for the volume at T °C, then

${\frac {V_{20}}{273.15+20}}={\frac {V_{0}}{273.15+0}}\quad \Longrightarrow V_{0}=273.15\times {\frac {24.0550}{293.15}}=22.4139\;\;[\mathrm {L} ]$

Problem 3

A certain amount of gas that has an initial pressure of 1 atm and an initial volume of 2 L, is compressed to a final pressure of 5 atm at constant temperature. What is the final volume of the gas?

Boyle's law (pV is constant)

$(1.1)\qquad \qquad p_{\mathrm {i} }\,V_{\mathrm {i} }=p_{\mathrm {f} }\,V_{\mathrm {f} }$

or

$(1.2)\qquad \qquad V_{\mathrm {f} }={\frac {p_{\mathrm {i} }\;V_{\mathrm {i} }}{p_{\mathrm {f} }}}$

Inserting the given numbers

$(1.3)\qquad \qquad V_{\mathrm {f} }=\left({\frac {1\cdot 2}{5}}\right)\;\left[{\frac {\mathrm {atm} \cdot \mathrm {L} }{\mathrm {atm} }}\right]=0.4\;[\mathrm {L} ]$

Ideal gas law

The number n of moles is constant

$(1.4)\qquad \qquad pV=nRT\quad \Longrightarrow \quad n={\frac {p_{\mathrm {i} }\,V_{\mathrm {i} }}{RT_{\mathrm {i} }}}={\frac {p_{\mathrm {f} }\,V_{\mathrm {f} }}{RT_{\mathrm {f} }}}$

It is given that the initial and final temperature are equal, $T_{\mathrm {i} }=T_{\mathrm {f} }\,$, therefore the products RT on both sides of the equation cancel, and Eq. (1.4) reduces to Eq. (1.1).

Problem 4

How many moles of nitrogen are present in a 50 L tank at 25 °C when the pressure is 10 atm? Numbers include only 3 significant figures.

$n={\frac {p\,V}{R\,T}}={\frac {10.0\cdot 50.0}{0.0821\cdot (273+25.0)}}\quad \left[{\frac {\mathrm {atm} \cdot \mathrm {L} }{{\frac {\mathrm {atm} \cdot \mathrm {L} }{\mathrm {K} \cdot \mathrm {mol} }}\cdot \mathrm {K} }}\right]={\frac {500}{0.0821\cdot 298}}\quad \left[{\frac {\mathrm {mol} \cdot \mathrm {atm} \cdot \mathrm {L} }{\mathrm {atm} \cdot \mathrm {L} }}\right]=20.4\quad [\mathrm {mol} ]$

Problem 5

Given is that dry air consists of 78.1% N₂, 20.1% O₂, and 0.8% Ar (volume percentages). The atomic weights of N, O, and Ar are 14.0, 16.0 and 39.9, respectively. Compute the mass of 1 m³ of dry air at 1 atm and 20 °C.

Answer

Since for ideal gases the volume V is proportional to the number of moles n, a volume percentage is equal to a molar percentage. For instance, for a mixture of two gases, it is easily shown that

${\frac {n_{1}}{n_{1}+n_{2}}}={\frac {V_{1}}{V_{1}+V_{2}}}$

which states that the molar percentage of gas 1 is equal to the volume percentage of gas 1.

The mass of 1 mole of dry air is

M = 0.781×28.0 + 0.201×32.0 + 0.008×39.9 g = 28.6192 g

In problem 1 it is found that the volume of 1 mole of ideal gas at 1 atm and 20 °C is 24.0550 L = 24.0550×10⁻³ m³, or

1 m³ contains 1/(24.0550×10⁻³) = 41.5714 mol

Hence the mass of 1 cubic meter of dry air is

M = 28.6192 × 41.5714 = 1189.7 g = 1.1897 kg