Answer
$Pi_{He}$=0.201 atm
Work Step by Step
Data
$n_{He}$ = 0.0410
The volume of vessel = 5L = $5\times 10^{-3}m^3$
Temperature = $25^{\circ}$C = 25 +273 = 298 K
$Pi_{He}$ = $\frac{n_{He}RT}{V}$
=$\frac{0.0410\times 8.314\times 298}{5\times 10^{-3}}$
=20316.09 $\frac{N}{m^{2}}$
$Pi_{He}$=0.201 atm
