Question

There is a gas cylinder filled with oxygen in front of you. It is 1.5 meters tall and 30 cm in diameter. The oxygen within the culinder has dropped to a gauge pressure of 2.52x10^6 Pa at a temperature of 20 C. This pressure is insufficient to ventilate a patient in respiratory distress. Youdo not have another oxygen cylinder present, and without oxygen the patient will die.

a. To maintain minimal ventilation, the pressure must be raised to a gauge pressure of 4.0x10^6 Pa. The only parameter you can vary is the temperature (by one of several means). What must the new temperature in the tank be to attain this minimal pressure?

b. What is the new absolute pressure in the tank?

c. How much gas is left in the tank? Give this value in two ways: the number of moles, and the more useful quantity of the number of Litres.

At this new pressure, the tank valve is opened and the oxygen is administered to the patient by nonbreather mask at a flow rate of 15 L/min leaving the tank.

d. How long do you have before the oxygen in the tnak is exhausted?

e. If the tank nozzle/outlet is a "Christmas Tree" fitting with an inside diameter of 0.4 cm, and the tube to the mask has an inside diameter of 0.5 cm, what is the average velocity of the gas just as it exits the "Christmas Tree" fitting?

Patm = 1.013x10^5 Pa

6.8912x10^3 Pa/psi

Density of O2 (approx.) = 1.3 kg/m^3

Oxygen= O2 = 32 g/mole

1 cm^3 = 1 ml

Viscosity of O2 (approx.) = 20.8x10^-6 Pa*s

Answer 1

Given that :

height of gas cylinder, h = 1.5 m

radius of gas cylinder, r = 15 cm = 0.15 m

initial gauge pressure, P₁ = 2.52 x 10⁶ Pa

initial absolute pressure, P_1,o = 26.2 x 10⁵ Pa

initial temperature, T₁ = 20 ⁰C = 293 K

(a) To maintain minimal ventilation, the pressure must be raised to a gauge pressure of 4 x 10⁶ Pa.

The new temperature in the tank to attain this minimal pressure which will be given as ;

using a gay-lussac law, we have

P_1,0 / T₁ = P_2,0 / T₂  

where, P_2,f = final absolute pressure = (1.013 x 10⁵ Pa) + (4 x 10⁶ Pa)

T₂ = (P_2,f / P_1,0) T₁

T₂ = [(41.0 x 10⁵ Pa) / (26.2 x 10⁵ Pa)] (293 K)

T₂ = (1.564) (293 K)

T₂ = 458.2 K

(b) The new absolute pressure in the tank which will be given as :

using an equation, P_abs = P_atmos + P_gauge

where, P_atmos = atmospheric pressure = 1.013 x 10⁵ Pa

P_gauge = new gauge pressure = 4 x 10⁶ Pa

the, we have

P_abs = (1.013 x 10⁵ Pa) + (4 x 10⁶ Pa)

P_abs = 4.1 x 10⁶ Pa

(c) number of moles of gas left in the tank which is given as :

using an ideal equation, we have

P_abs V = n R T₂

where, V = volume of the cylinder =r² h

V = (3.14) (0.15 m)² (1.5 m) = 0.106 m³

then, we have

(4.1 x 10⁶ Pa) (0.106 m³) = n (8.314 J/mol.K) (458.2 K)

(0.4346 x 10⁶ J) = n ( J/mol)

n = (0.4346 x 10⁶ J) / (3809.4 J/mol)

n = 114.1 moles

At this new pressure, the tank valve is opened and the oxygen is administered to the patient by nonbreather mask at a flow rate of 15 L/min leaving the tank.

(d) Time taken by the oxygen which is given as :

rate of oxygen flown out = 15 L/min

then, we get

t = (106 L) / (15 L/min)

t = 7 min

(e) The average velocity of the gas just as it exits the "Christmas Tree" fitting which will be given as :

using an equatuion, we have

Q = A v

where, A = area of cross-section = r²

Q = flow rate = 15 L/min = 15 x 10^-3 m³/min

then, we have

(15 x 10^-3 m³/min) = [(3.14) (2.5 x 10^-3 m)²] v

v = (15 x 10^-3 m³/min) / (19.6 x 10^-6 m²)

v = 0.7653 x 10³ m/min

v = 765.3 m/min

v = 12.7 m/s