(ઉપયોગ $R$ $=0.083\,L\,bar\,K ^{-1}\,mol ^{-1}$ )
$=\frac{1 \times 2}{60000 \times 0.2} \times 0.083 \times 300$
$=0.00415$ bar $\quad\left(\because 1\right.$ bar $\left.=10^{5} Pa \right)$
So, $0.00415 \times 10^{5} Pa =415 \,Pa$
$(R =0.083\, L\, bar \,K ^{-1} \,mol ^{-1})$ (નજીકનો પૂર્ણાંક)