91Ó°ÊÓ

Given data:

$\begin{array}{r}\mathrm{Temp}\left(T\right)=295K,\text{Length}\left(L\right)=1.5\mathrm{m},\mathrm{Pr}\text{essure}\left(P\right)=21\mathrm{atm}\\ \mathrm{MolarMass}\left(M\right)=32\mathrm{g}/\mathrm{mol},\text{Radius}\left(r\right)=0.45\mathrm{m}\end{array}$

Calculating the volumeof the cylinder for the volume of the oxygen

$\begin{array}{r}V=\mathrm{π}{r}^{2}L\\ =\mathrm{π}×(0.45{)}^2×1.5\\ =0.953{\mathrm{m}}^3\end{array}$

For the calculation of no. of moles(n)present in the cylinder,Ideal Gas Equation is applied

Here, Gas constant (R) = 8.315 J/mol

Pressure is in the Pascal unit therefore it is multiplied by $1.013×{10}^{5}Pa/atm$

localid="1668399953184" $\begin{array}{r}P.V=n.R.T\\ n=\frac{PV}{RT}\\ =\frac{21atm×1.013×{10}^5\mathrm{Pa}/\mathrm{atm}×0.953{\mathrm{m}}^3}{8.315\mathrm{J}/\mathrm{mol}×295\mathrm{K}}\\ =827\mathrm{mol}\end{array}$

Therefore, the number of moles of oxygen present in the canister is 827 mol

Mass of oxygen (m) present in the canister

$\begin{array}{r}m=n.M\\ =827×32×{10}^{−3}\mathrm{kg}/\mathrm{mol}\\ =26.5\mathrm{kg}\end{array}$

Hence, 26.5 kg of mass is increased due to the presence of oxygen in the canister.