The compressibility factor Z=( text PV / text nRT ) for hydrogen gas at 273 K and 1 atm pressure is

Question

The compressibility factor Z=( text PV / text nRT ) for hydrogen gas at 273 K and 1 atm pressure is (A) Zero (B) one (C) greater than one (D) between zero and one

Tardigrade Admin · Accepted Answer

At NTP condition (p=1 atm , T=273 K ) For 1 mole (n=1) of H2(g), V=22.4 L mol -1. So, Z=(p V/n R T)=(1 × 224/1 × 0.082 × 273) [∵ R=0.082 L. atm .m o l-1 K-1] =1.000625 ⇒ Z >1

Q. The compressibility factor $Z = \frac{PV}{nRT}$ for hydrogen gas at $273 K$ and $1 a t m$ pressure is

Zero

one

greater than one

between zero and one

At NTP condition $(p = 1 a t m, T = 273 K)$
For $1$ mole $(n = 1)$ of $H_{2} (g), V = 22.4 L m o l^{- 1}$ .
So, $Z = \frac{p V}{n RT} = \frac{1 \times 224}{1 \times 0.082 \times 273}$
$[∵ R = 0.082 L$ atm $m o l^{- 1} K^{- 1}]$
$= 1.000625$
$\Rightarrow Z > 1$

Q. The compressibility factor Z= nRT PV ​ for hydrogen gas at 273K and 1atm pressure is

Zero

one

greater than one

between zero and one

At NTP condition (p=1atm,T=273K) For 1 mole (n=1) of H2​(g),V=22.4Lmol−1. So, Z=nRTpV​=1×0.082×2731×224​ [∵R=0.082L atm mol−1K−1] =1.000625 ⇒Z>1

Q. The compressibility factor $Z = \frac{PV}{nRT}$ for hydrogen gas at $273 K$ and $1 a t m$ pressure is

At NTP condition $(p = 1 a t m, T = 273 K)$
For $1$ mole $(n = 1)$ of $H_{2} (g), V = 22.4 L m o l^{- 1}$ .
So, $Z = \frac{p V}{n RT} = \frac{1 \times 224}{1 \times 0.082 \times 273}$
$[∵ R = 0.082 L$ atm $m o l^{- 1} K^{- 1}]$
$= 1.000625$
$\Rightarrow Z > 1$