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 $${22.4\ \mathrm L}$$?

1 Answer

The issue is that one mole of a gas at NTP should not be 22.4 L based on the ideal gas. This is only true for a gas at STP. The key difference between the two conditions is they are defined by different temperatures (STP is for 273 K while NTP is for 293 K).

They are also different in that the IUPAC now defines STP with respect to 1 bar while NTP is defined with respect to 1 atm.

See The Engineering ToolBox for more info.

• 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 '17 at 2:37
• @PrittBalagopal good point, I will make note of that explicitly. Jun 15 '17 at 2:54
• You should use 298K for NTP! Jun 15 '17 at 8:15
• @Vaibhav Dixit all the definitions I have seen for NTP define it at 20C or 293K. Jun 15 '17 at 19:52