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 ${22.4\ \mathrm L}$?
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. 

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