One gram mole of a gas at NTP occupies 22.4 L. This fact was derived from

Question

One gram mole of a gas at NTP occupies 22.4 L. This fact was derived from (A) law of gaseous volumes (B) Avogadro's hypothesis (C) Berzelius hypothesis (D) Dalton's atomic theory

Tardigrade Admin · Accepted Answer

According to ideal gas equation, P V=n R T Putting the standard conditions, i.e. P =1 atm n=1 mole, R=0.0821 L atm K-1, mol -1, T=273 K V=(n R T/P)=(1 text mole × 0.0821 L atm K-1 mol -1 × 273 K /1 atm ) V=22.4

Q. One gram mole of a gas at NTP occupies $22.4 L$ . This fact was derived from

law of gaseous volumes

Avogadro's hypothesis

Berzelius hypothesis

Dalton's atomic theory

According to ideal gas equation, $P V = n RT$
Putting the standard conditions,
i.e. $P = 1$ atm $n = 1$ mole,
$R = 0.0821 L$ atm $K^{- 1}$ , mol $^{- 1},$
$T = 273 K$
$V = \frac{n RT}{P} = \frac{1 mole \times 0.0821 L a t m K ^{- 1} m o l ^{- 1} \times 273 K}{1 a t m}$
$V = 22.4$

Q. One gram mole of a gas at NTP occupies 22.4L. This fact was derived from