Question

An empty cylindrical canister 1.50 m long and 90.0 cm in diameter is to be filled with pure oxygen at 22.0°C to store in a space station. To hold as much gas as possible, the absolute pressure of the oxygen will be 21.0 atm. The molar mass of oxygen is 32.0 g>mol. (a) How many moles of oxygen does this canister hold? (b) For someone lifting this canister, by how many kilograms does this gas increase the mass to be lifted?

Short answer

(a) Moles of oxygen contained in the canister are 827 mol.

(b) 26.5 Kg of mass increased.

Step-by-step solution

(a) No. of moles of Oxygen in the canister

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

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

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 $1.013\times {10}^{5}Pa/atm$

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

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

(b) Increase in mass due to gas present in the canister

Mass of oxygen (m) present in the canister

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

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