Published by Patrick Mutisya · 14 days ago
Understand that a gas which obeys the relationship \$pV \propto T\$, where \$T\$ is the thermodynamic temperature, is called an ideal gas.
If \$pV\$ is proportional to \$T\$, we can write
\$pV = k\,T\$
where \$k\$ is a constant for a given amount of gas.
When the amount of gas is expressed in moles (\$n\$), the constant becomes \$nR\$, where \$R\$ is the universal gas constant (\$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}\$). Thus the equation becomes
\$pV = nRT\$
This is the familiar ideal‑gas law.
| Symbol | Quantity | SI Unit | Typical Symbol in Equations |
|---|---|---|---|
| \$p\$ | Pressure | pascal (Pa) | p |
| \$V\$ | Volume | cubic metre (m³) | V |
| \$T\$ | Thermodynamic temperature | kelvin (K) | T |
| \$n\$ | Amount of substance | mole (mol) | n |
| \$R\$ | Universal gas constant | J mol⁻¹ K⁻¹ | R |
Calculate the pressure exerted by 2.00 mol of an ideal gas occupying a volume of 0.050 m³ at a temperature of 300 K.
Using \$pV = nRT\$:
\$p = \frac{nRT}{V} = \frac{(2.00\ \text{mol})(8.314\ \text{J mol}^{-1}\text{K}^{-1})(300\ \text{K})}{0.050\ \text{m}^3}\$
\$p = \frac{4988.4\ \text{J}}{0.050\ \text{m}^3} = 9.98\times10^{4}\ \text{Pa}\$
Thus the pressure is approximately \$1.0\times10^{5}\ \text{Pa}\$ (or 1 atm).
A gas that satisfies the proportionality \$pV \propto T\$ for a fixed amount of substance follows the ideal‑gas law \$pV = nRT\$. This simple relationship is a cornerstone of thermodynamics and provides a useful first approximation for the behaviour of many gases under ordinary conditions.