Question

A $500-cc$ bulb weights $38.734$ grams when evacuated and $39.3135$ grams when filled with air at $1atm$ pressure and ${24}^{o}C$. Assuming that air behaves as an ideal gas at this pressure, calculate effective mass of $1$ mole of air.

Answer 1

Wt of empty vessel $=38.734g$

Wt. of filled vessel $=39.3135g$

$\therefore$ Wt of air filled $=39.3135-38.734=0.5795g$

Volume of bulb $=500cc=5\times {10}^{-1}lit$

$T={24}^{o}C=(273+24)K=297K$

Effective mass of $1mole$ of air =Mol. wt of air

By, $PV=nRT$

$1\times 5\times {10}^{-1}=\frac{0.5795}{M.W}\times 0.0821\times 297$

$M.W.=\frac{0.5795\times 0.0821\times 297}{0.5}$

$\frac{0.5795\times 0.0821\times 297}{0.5}=28.26g$