8.3: Ideal Gas Law

In most materials, if we know three of the four thermodynamic properties, volume, pressure, temperature, and entropy, we can derive the fourth property as well as other thermodynamic measures. Such materials are called simple compressible systems [109, 102]. For such materials, the ideal gas law relates pressure, volume, and temperature.

$$\mathbb{P}\mathbb{V} = \mathbb{N}\mathbb{R}T. \nonumber$$

While this is not a mathematical law, it is a good description of gases, and it can be used as a rough approximation for liquids and solids. Consider a container filled with a gas. If the volume of the container is compressed while the temperature is kept constant, the pressure increases. If the gas is heated and the pressure is kept constant, the volume increases. The energy stored in a gas that undergoes change in volume at constant temperature is given by

$$E = \int \mathbb{P} d \mathbb{V} \nonumber$$

where the change in energy is specified by

$$\Delta E = \mathbb{P} \Delta \mathbb{V}. \nonumber$$

The ideal gas law can be written incorporating entropy as

$$\mathbb{P} \mathbb{V} = ST. \nonumber$$

For example, consider a 10 L tank that holds 5 mol of argon atoms. The argon gas is at a temperature of $T = 15 ^{\circ}C$. Find the pressure in the tank in pascals and in atm. We begin by converting the volume and temperature to more convenient units, $\mathbb{V} = 0.01 m^3$ and $T = 288.15$ K. Next, the ideal gas law provides the pressure in Pa.

$$\mathbb{P} =\frac{\mathbb{N} \mathbb{R}\mathrm{T}}{\mathbb{V}} =\frac{5 \mathrm{mol} \cdot 8.314 \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}} \cdot 288.15 \mathrm{K}}{0.01 \mathrm{m}^3} =1.20 \cdot 10^6 \mathrm{Pa} \nonumber$$

Finally, we convert the pressure to the desired units.

$$\mathbb{P} =1.20 \cdot 10^{6} \mathrm{Pa} \cdot \frac{1 \mathrm{atm}}{101325 \mathrm{Pa}} =11.8 \mathrm{atm} \nonumber$$

As another example, consider a container that holds neon atoms at a temperature of $T = 25 ^{\circ}C$. Assume that the pressure in the container is 10 kPa, and the mass of the neon in the container is 3000 g. Find the volume of the container. The temperature is 298.15 K. From a periodic table, the atomic weight of a neon atom is $20.18 \frac{g}{mol}$. Thus, the container holds 148.7 mol of neon atoms. Next, we use the ideal gas law to find the volume.

$$\mathrm{V} =\frac{\mathbb{N} \mathbb{R}\mathrm{T}}{\mathbb{P}} =\frac{148.7 \mathrm{mol} \cdot 8.314 \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}} \cdot 298.15 \mathrm{K}}{10^{4} \mathrm{Pa}}=36.86 \mathrm{m}^{3} \nonumber$$