An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressures at temperature 27°C. The mass of the oxygen withdrawn from the cylinder is nearly equal to:

[Given, R= 100/12 Jmol^-1 K ^-1, and molecular mass of O₂ = 32, 1 atm pressure = 1.01 × 10⁵ N/m]

The problem gives the initial amount of oxygen in the cylinder (18.20 mol) and asks for the mass that is withdrawn when the gauge pressure falls to 11 atm at 27 °C.

First convert the temperature to kelvin:

$T\; =\; 27^\backslash circ\backslash text\{C\}\; +\; 273\; =\; 300\backslash \; \backslash text\{K\}$

Gauge pressure of 11 atm means the absolute pressure is one atmosphere higher:

$P\_\{\backslash text\{abs\}\}\; =\; (11\; +\; 1)\backslash \; \backslash text\{atm\}\; =\; 12\backslash \; \backslash text\{atm\}$

Convert the pressure to SI units (1 atm = 1.01 × 10⁵ N m^–2):

$P\; =\; 12\; \backslash times\; 1.01\; \backslash times\; 10^\{5\}\backslash \; \backslash text\{N m\}^\{-2\}\; =\; 1.212\; \backslash times\; 10^\{6\}\backslash \; \backslash text\{N m\}^\{-2\}$

The cylinder volume is 30 L, which equals 0.030 m³:

$V\; =\; 30\backslash \; \backslash text\{L\}\; =\; 30\; \backslash times\; 10^\{-3\}\backslash \; \backslash text\{m\}^\{3\}\; =\; 0.030\backslash \; \backslash text\{m\}^\{3\}$

Use the ideal‑gas equation to find the number of moles remaining after the withdrawal:

$P\; V\; =\; n\_\{\backslash text\{final\}\}\; R\; T$

Given R = 100/12 J mol^–1K^–1 (≈ 8.33 J mol^–1K^–1), solve for $n\_\{\backslash text\{final\}\}$:

$n\_\{\backslash text\{final\}\}\; =\; \backslash frac\{P\; V\}\{R\; T\}\; =\; \backslash frac\{(1.212\; \backslash times\; 10^\{6\})(0.030)\}\{(100/12)(300)\}$

Calculate the denominator:

$(100/12)(300)\; =\; \backslash frac\{100\; \backslash times\; 300\}\{12\}\; =\; \backslash frac\{30000\}\{12\}\; =\; 2500\backslash \; \backslash text\{J mol\}^\{-1\}$

Calculate the numerator:

$P\; V\; =\; 1.212\; \backslash times\; 10^\{6\}\; \backslash times\; 0.030\; =\; 36360\backslash \; \backslash text\{J\}$

Thus

$n\_\{\backslash text\{final\}\}\; =\; \backslash frac\{36360\}\{2500\}\; =\; 14.544\backslash \; \backslash text\{mol\}$

The amount of oxygen withdrawn is the difference between the initial and final moles:

$\backslash Delta\; n\; =\; n\_\{\backslash text\{initial\}\}\; -\; n\_\{\backslash text\{final\}\}\; =\; 18.20\backslash \; \backslash text\{mol\}\; -\; 14.544\backslash \; \backslash text\{mol\}\; =\; 3.656\backslash \; \backslash text\{mol\}$

Convert the withdrawn moles to mass using the molar mass of O₂ (32 g mol^–1 = 0.032 kg mol^–1):

$m\; =\; \backslash Delta\; n\; \backslash times\; M\; =\; 3.656\backslash \; \backslash text\{mol\}\; \backslash times\; 0.032\backslash \; \backslash text\{kg mol\}^\{-1\}\; =\; 0.1170\backslash \; \backslash text\{kg\}$

Rounded to three significant figures, the mass withdrawn is

$m\; \backslash approx\; 0.116\backslash \; \backslash text\{kg\}$

Therefore, the mass of oxygen withdrawn from the cylinder is approximately 0.116 kg, which matches the given correct option.