A 10 ft^3 tank contains gas at a pressure of 500 psia, temperature of 85 Fahrenheit and a weight of 25 lbs. A part of the gas was discharged and the temperature and pressure changed to 70 fahrenheit and 300 psia, respectively. Heat was applied and the temperature was back to 85 fahrenheit. Find the final weight, volume, and pressure of the gas.
Answer : 15.43 lb; 10 ft^3; 308.5 psia
Answer : 15.43 lb; 10 ft^3; 308.5 psia
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Here is the method.
Convert temperatures (T1=85degF and T2=70degF) to centigrade, then add 273 to convert to kelvin. Use the kelvin temperatures throughout.
Since PV = nRT
PV/(nT) = R
Providing we are consistent with units and have T in kelvin:
P1V1/(n1T1) = P2V2/(n2T2) where
P is pressure in psia
V is volume in ft^3
n is weight (proportional to mass and moles) in lb
T is in kelvin
The volume is fixed - it remains 10ft^3 throughout.
For the discharge process:
P1V1/(n1T1) = P2V2/(n2T2)
500x10/(25xT1) = 300x10/(n2xT2)
5/(25xT1) = 3/(n2xT2)
This allows the weight of gas, n2, to be found.
For the final stage:
P1V1/(n1T1) = P2V2/(n2T2)
300x10/(n2xT2) = P2x10/(n2xT1)
300/T2 = P2/T1
This allowed P2 to be found.
Convert temperatures (T1=85degF and T2=70degF) to centigrade, then add 273 to convert to kelvin. Use the kelvin temperatures throughout.
Since PV = nRT
PV/(nT) = R
Providing we are consistent with units and have T in kelvin:
P1V1/(n1T1) = P2V2/(n2T2) where
P is pressure in psia
V is volume in ft^3
n is weight (proportional to mass and moles) in lb
T is in kelvin
The volume is fixed - it remains 10ft^3 throughout.
For the discharge process:
P1V1/(n1T1) = P2V2/(n2T2)
500x10/(25xT1) = 300x10/(n2xT2)
5/(25xT1) = 3/(n2xT2)
This allows the weight of gas, n2, to be found.
For the final stage:
P1V1/(n1T1) = P2V2/(n2T2)
300x10/(n2xT2) = P2x10/(n2xT1)
300/T2 = P2/T1
This allowed P2 to be found.
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Discuss in my forum. Ideal Gas Law - Worked Chemistry Problems . Example Problem #1. By Anne Marie Helmenstine, Ph.D., About.com Guide.