# Proving using ideal gas law that molar volume at NTP is 22.4 litres

The definition of NTP as given by IUPAC is

NTP – Normal Temperature and Pressure – is defined as air at $20\ ^\circ\mathrm C$ ($\pu{293.15 K}$, $\pu{68^{\circ}F}$) and $\pu{1 atm}$ ($\pu{101.325 kN/m2}$, $101.325\ \mathrm{kPa}$, $14.7\ \mathrm{psia}$, $0\ \mathrm{psig}$, $29.92\ \mathrm{inHg}$, $407\ \mathrm{in}\ce{H2O}$, $760\ \mathrm{Torr}$).

It is also known that volume occupied by $1$ mole of gas at NTP is $\pu{22.4 L}$.

My attempt at proving the above statement,

For $1$ mole, $PV=RT$

Substituting $P=\pu{1 atm}$, $T=\pu{293 K}$, $R=0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$, we get $V=\pu{24.05 L}$ as the molar volume.

On putting $T=\pu{273 K}$ and $P=\pu{1 bar}$ (conditions of STP), we get $V=\pu{22.2 L}$

Therefore, molar volume at STP is $\pu{22.4L}$.

Why is the above calculation in NTP condition giving wrong results? How else to prove that molar volume at NTP is ${22.4\ \mathrm L}$?

• And to add further point, the pressure is one bar, not one atm. So at STP it would be more correctly $\pu{22.7L}$. – Pritt says Reinstate Monica Jun 15 '17 at 2:37