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HISTORY OF CIVIL ENGINEER(ING)
Typology: Cheat Sheet
Uploaded on 11/11/2024
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Ideal Gas
Ideal Gas, defined
Equation of State PV = nRT where: P = absolute pressure R = universal gas constant V = volume T = absolute temperature (K, deg R) n = number of moles
Derived from PV = nRT
n = m/M
Standard Temperature and Pressure (STP) standard temperature = 273 K = 492 Ro^ = 0 Co^ =32 Fo standard pressure = 1 atm = 14.7 psi = 760 mm Hg = 760 torr = 101.325 kPa = 29.92 in Hg
Molar volume of any ideal gas at STP V = 22.4 L/mole
Universal Gas Constant
R = 0 .0821
L atm gmol K
gmol K
3 o
ft atm lbmol R
n = m/M
R = PV/nT 1 kgmol = 1000 gmol = 2.2 lbmol
Combined Gas Law BT CP GV PV = nRT PV/T = nR nR = constant n 1 = n 2
(^1 1) = 2 2 1 2
subscript 1 = initial condition subscript 2 = final condition
Boyle’s Law nRT = constant P V 1 1 =P V 2 2
Charles’ Law P = constant = nR (^1) = 2 1 2
Gay-Lussac’s Law V = constant = nR (^1) = 2 1 2
Example 1 A sample of gas occupies a volume of 225 ml at a pressure of 720 mm Hg and a temperature of 20oC. Calculate the new pressure if the volume is increased to 350 ml at constant temperature.
2 2
P 463 mm Hg
Example 2 A sample of gas occupies a volume of 275 ml at 20oC and 1 atm pressure. Calculate the volume if the volume of the gas at 0^0 C and 1 atm pressure.
2
2
V 256 ml
An ideal gas at 760 mm Hg (absolute) is cooled from 40 deg F to 80 deg F. What is the new absolute pressure of this gas in mm Hg?
10 points 1520
380 none of the above What is the volume of a 1.25-mol gas at 1 atm absolute and 20 deg C? Assume the gas behaves like an ideal gas. R = 0.0821 L-atm/mol-K.
10 points
none of the above The absolute pressure of a 1.95-liter gas is increased from 101.3 kPa to 110 kPa. Determine the new volume.
10 points
none of the above A gas at 14.7 psia is heated from 300 K to 371 K. Find the new pressure in psia?
10 points
none of the above