# The Gas Equation

If Boyle’s and Charle’s law are combined, then

$V \propto \dfrac{T}{P} \text{or} PV \propto T$

PV = RT (R = gas constant)

$\dfrac{PV}{T} = \text{constant} R$

Or PV = nRT

Or $\dfrac{P_1V_1}{T_1} = \dfrac{P_1V_2}{T_2}$

Gas equation PV = RT is for one mole of gas. For n moles of gas this equation is nRT, and is known as Ideal Gas Equation. (n = m/M, m =mass of the gas and M =molar mass of the gas).

Nature of R:

From gas equation:

$R = \dfrac{PV}{T} = \dfrac{\text{Pressure} \times \text{Volume}}{\text{Temperature}} \\[3mm] R = \dfrac{(\text{ Force}) (\text{Length})}{(\text{Temperature})} = \dfrac{\text{Work}}{\text{Temperature}}$

Thus, R represents work done per degree per mole and is expressed in $JK mol^{-1}$ (in SI units).

Value of R:

The value of R may be calculated by substituting the values of P, V and T in gas equation.

In SI units, p = $1.03125 \times 10^5 Nm^{-2}, V = 22400 \times 10^{-6}m^3 \text{and} T = 273K$, Hence

$R = \dfrac{PV}{T} \\[3mm] \dfrac{(1.03125 \times 10^5 Nm^{-2})(22400 \times 10^{-6}m^3)}{(273K)} \\[3mm] = 8.31jk^{-1}MOL^{-1} (1j = 10^7 erg \\[3mm] 8.314 \times 10^7 erg K^{-1} mol^{-1} \\[3mm] 8.314 \times 10^7 dyne cm K^{-1} mol^{-1} (erg = dyne \times cm) \\[3mm] 1.99 cal K^{-1} (1J = 0.2390 cal) \\[3mm] \text{If} P = 1 atm, V= 22.4 L mol^{-1} \text{and} T = 273 K. \\[3mm] \text{then}, \\[3mm] R = \dfrac{PV}{T} \\[3mm] \dfrac{(1atm)(22.4L mol^{-1})}{(273 K)} \\[3mm] 0.0821 atm K^{-1} mol^{-1}$

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