Turns out my assumptions were wrong about the stated 9450 L of NX2 gas. The 9450 L is the total volume of standard pressure gas compressed into the cylinder to 20 MPa. Using V2=V1/(p2/p1) having p1=0.101 MPa, p2=20 MPa, V1=9450 L. The total volume of gas within the cylinder is about 47 L a reasonably fitting size for the cylinder dimensions. Hence to answer my own question, at the pressure of usage (0.45 MPa) the volume of gas (using same equation as above) is 2121 L which at 60 cL/min equates to 59 h of flow roughly.

Final answers: (i) The number of moles of gas in the cylinder is approximately ( 166.67 , moles ). (ii) The number of gas atoms in the cylinder is approximately ( 1.003 * 10^(26) , atoms ).. To solve the problem, we will use the Ideal Gas Law, which is given by the equation: [ PV = nRT ] where: - ( P ) = pressure (in Pascals) - ( V ) = volume (in cubic meters) - ( n ) = number of moles of gas - ( R ) = ideal gas constant (( 8.314 , J/(mol·K) )) - ( T ) = temperature (in Kelvin) **Step 1:** Convert the temperature from Celsius to Kelvin. [ T(K) = T(°C) + 273.15 = 17 + 273.15 = 290.15 , K ] **Step 2:** Convert the pressure from MPa to Pascals. [ P = 20 , MPa = 20 * 10^(6 , Pa) = 2.0 * 10^(7 , Pa) ] **Step 3:** Substitute the values into the Ideal Gas Law to find the number of moles ( n ). [ n = PV/RT = frac(2.0 * 10^(7 , Pa))(2.0 * 10^(-2) , m^((3))(8.314 , J/(mol·K)))(290.15 , K) ] **Step 4:** Calculate ( n ). [ n = frac(2.0 * 10^(7)(2.0 * 10^-2))(8.314)(290.15) = frac4.0 * 10^52410.59 approx 166.67 , moles ] **Step 5:** Now, to find the number of gas atoms in the cylinder, we will use Avogadro's number ( N_A ), which is approximately ( 6.022 * 10^(23) , atoms/mol ). **Step 6:** Calculate the number of gas atoms. [ Number of atoms = n * N_A = 166.67 , moles * 6.022 * 10^(23) , atoms/mol ] **Step 7:** Calculate the total number of atoms. [ Number of atoms approx 166.67 * 6.022 * 10^(23) approx 1.003 * 10^(26) , atoms ]

"Concept: By thin-walled cylinder, we mean that the thickness ‘t' is very much smaller than the radius Ri and we may quantify this by stating that the ratio t Ri of the thickness of radius should be less than 0.1. In a thin shell circumferential stress is ({sigma _c} = frac{{Pd}}{{2t}};) and Longitudinal stress will be half of the circumferential stress i.e. ({sigma _l} = frac{{Pd}}{{4t}}). where p is the pressure, t is the thickness of cylinder, d is the diameter of the cylinder Calculation: Given: p = 20 MPa, t = 2.5 mm and d = 50 mm. Longitudinal stress ({sigma _l} = frac{{Pd}}{{4t}}) ({sigma _l} = frac{{20times 50}}{{4times 2.5}} =100~MPa) Hoop stress ({sigma _h} = frac{{Pd}}{{2t}} = {sigma _1}) Hoop strain ({epsilon_h} = frac{{Pd}}{{4tE}}left( {2 - mu } right)) Longitudinal stress ({sigma _L} = frac{{Pd}}{{4t}} = {sigma _2} = frac{{{sigma _1}}}{2}) Longitudinal strain ({epsilon_L} = frac{{Pd}}{{4tE}}left( {1 - 2mu } right) = frac{{{sigma _2}}}{E}left( {1 - 2mu } …

Commercial gas cylinders often come at an initial pressure of 220bar, but all the vendor/ look at the data sheet. So yes. Very possible Remember - More details = better answers Also: If you get a response it's polite to respond to it.