Question

What is the mass of $2L$ of nitrogen at $22.4arm$ pressure and $273K$ ?

• A. $28g$
• B. $14\times 22.4g$
• C. $56g$
• D. None of these

Answer 1

Answer: (A)

From ideal gas equation
$pV=nRT$
where $p$ is pressure,
$V$ the volume,
$R$ the gas constant,
$T$ the temperature and
$n$ the number of moles.
$\therefore n=\frac{pV}{RT}$
Given $p=22.4atm$ pressure
$=22.4\times 1.01\times 10^{5}N{m}^{-2}$
$V=2L=2\times 10^{-3}{m}^3$
$R=8.31Jmo{l}^{-1}-K^{-1}$
$T=273K$
$\therefore n=\frac{22.4\times 1.01\times 10^5\times 2\times 10^{-3}}{8.31\times 273}$
$n=1.99\approx 2$
$\therefore n=\frac{\text{ mass }}{\text{ atomic weigh }}$
We have mass $=n\times$ atomic weight
$=2\times 14=28g$