Question

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

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

Answer 1

Answer: (A)

$28\,g$

Answer 2

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.
$\therefore n=\frac{p V}{R T}$
Given $p =22.4\, atm$ pressure
$=22.4 \times 1.01 \times 10^{5} Nm ^{-2}$
$V=2 L=2 \times 10^{-3} m ^{3}$
$R=8.31 \,J\,mol ^{-1}- K ^{-1}$
$T =273 \,K$
$\therefore n=\frac{22.4 \times 1.01 \times 10^{5} \times 2 \times 10^{-3}}{8.31 \times 273}$
$n=1.99 \approx 2$
$\therefore n=\frac{\text { mass }}{\text { atomic weigh }}$
We have mass $=n \times$ atomic weight
$=2 \times 14=28 \,g$