The → partial pressure of a gaseous substance $X$ in a gas mixture of volume $V$ is its pressure supposing that it occupies alone the whole volume .
The ideal gas law allows to calculate the partial pressure of a gas $X$ in a gas mixture of volume $V$: $P_X$ $=$ $\frac{n_X\cdot R\cdot T }{V}$ $=$ $\frac{n_X}{V}\cdot R\cdot T$
Partial pressure of a gas $X$ in a gas mixture of volume $V$: $P_X$ $=$ $[X]\cdot R\cdot T$ (1) where $[X]$ is the molarity (molar concentration) of the gas $X$ in the gas mixture.
As gaseous equilibria $aX$ $+$ $bY$ $...$ $\leftrightarrows$ $cC$ $+$ $dD$ $... $ are always proceding in one single phase, the law of mass action applies: $K_c$ $=$ $\frac{[C]^c[D]^d...}{[X]^a[Y]^b...}$ (2) Introducing (1) into (2): $K_c$ $=$ $\frac{(\frac{P_C}{RT})^c(\frac{P_D}{RT})^d...}{(\frac{P_X}{RT})^a(\frac{P_Y}{RT})^b...}$ $=$ $\frac{P_C^cP_D^d...}{P_X^aP_Y^b...}(RT)^{a+b+...-(c+d+.. )}$ As $K_c$ depends only on the temperature, this applies also to : $K_p$ $=$ $\frac{K_c}{(RT)^{a+b+...-(c+d+.. )}}$ and so we have:
Equilibrium between gases: $K_p$ $=$ $\frac{P_C^cP_D^d...}{P_X^aP_Y^b...}$ where $K_p$ is the pressure equilibrium constant.
As a result of the previous calculation, we have :
$K_c$ $=$ $K_p(RT)^{a+b+...-(c+d+..)}$ where $a$ $+$ $b$ $+$ $... $ $-(c+d+.. )$ is the difference between the sum of coefficients of the reagents and the sum of coefficients of the products as shown in the equation.