# 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 $\pu{22.7L}$. Jun 15, 2017 at 2:37