Many gases are shipped in high-pressure containers. consider a steel tank whose volume is 55.0 gallons that contains o2 gas at a pressure of 16,500 kpa at 23 °c. (a) what mass of o2 does the tank contain? (b) what volume would the gas occupy at stp? (c) at what temperature would the pressure in the tank equal 150.0 atm? (d) what would be the pressure of the gas, in kpa, if it were transferred to a container at 24 °c whose volume is 55.0 l?
V = 55.0 gallons x 3.78541 = 208 L
P = 16500 kPa
T = 23 + 273.15 = 296.15 K
Part A) From the equation PV =
mass of O₂ = (16500 kPa x 208 L x 32 g/mol) / (8.314 x 296.15 K)
= 44.6 Kg
Part B) at STP we have:
T = 273.15 K and P = 101.3 kPa
so from PV = mRT / M
V = (44600 x 8.314 x 273.15) / (32 x 101.3)
= 31248 L
Part C) From the equation PV = mRT / M
T = (150 atm x 101.3 kPa / atm x 208L x 32g/mol) / (8.314 x 44600)
= 272.8 K
Part D) we have:
T = 24 + 273.15 K = 297.15 K
so from PV = mRT / M, we get
P = (44600 x 8.314 x 297.15) / (32 x 55)
= 62464 kPa
To find the mass, you need to find how many moles of oxygen did the tank contain using ideal gas formula. Remember to change the unit into Kelvin and Liters.
n= 16,500 kpa (55 gallons *3.78541 liters/gallon) / 8.314 * (23+273.15)K
weight= n * molecular weight= 1395.20* 18= 25113.7 gram
(b) what volume would the gas occupy at stp?
STP condition defined as 273K temperature and 1atm pressure. Ideal gas will have 22.4L/mol volume in STP.
volume in STP= n * 22.4L/mol= 1395.20 moles* 22.4L/mol= 31252.48 L
(c) at what temperature would the pressure in the tank equal 150.0 atm?
Since this question using atm as the pressure unit, you have to change the ideal constant into R=0.08205 L atm / mol·K
T= 150.0 atm * 208.19L / 1395.20mol *0.08205 L atm / mol·K
T= 272.79 K
(d) what would be the pressure of the gas, in kpa, if it were transferred to a container at 24 °c whose volume is 55.0 l?
The volume and temperature are changed and you are asked to calculate the pressure. You can use the same formula but modified it to calculate the pressure.
P= 1395.20mol * 8.314 * (24+273.15) / 55 L
P= 62,669.97 kpa
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