Learning Goal: To be able to predict and calculate properties of real gases. Kin

Question

Learning Goal:

To be able to predict and calculate properties of real gases.

Kinetic molecular theory makes certain assumptions about gases that are, in fact, not true for real gases. Therefore, the measured properties of a gas are often slightly different from the values predicted by the ideal gas law. The van der Waals equation is a more exact way of calculating properties of real gases. The formula includes two constants, a and b, that are unique for each gas.

The van der Waals equation is (P+an2V2)(V?nb)=nRT, where P is the pressure, n the number of moles of gas, V the volume, T the temperature, and R the gas constant.

All real gases possess intermolecular forces that can slightly decrease the observed pressure. In the van der Waals equation, the variable P is adjusted to P+an2V2, where a is the attractive force between molecules.

Part A The van der Waals Equation In general, which of the following gases would you expect to behave the most ideally even under extreme conditions? Learning Goal: To be able to predict and calculate properties of real gases. O CO Kinetic molecular theory makes certain assumptions about gases that are, in fact, not true for real gases. Therefore, the measured properties of a gas are often slightly different from the values predicted by the ideal gas law. The van der Waals Submit Hints My Answers Give Up Review Part uation is a more exact way of calculating properties of real gases. The formula includes two constants, a and b, that are unique for each gas. Part B In general, which of the following gases would you expect to behave the least ideally under extreme conditions? The van der Waals equation is (PA ) (V-nb) = nRT, where ) (V-nb) = nRT, where P is the P + essure, n the number of moles of gas, V the volume, T the temperature, and R the gas constant. O CO All real gases possess intermolecular forces that can slightly decrease the observed pressure. In the van der Waals equation, the variable P is adjusted to P where a is the attractive force between molecules Submit Hints My Answers Give Up Review Part Part C 12.0 moles of gas are in a 8.00 L tank at 20.9 °C . Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a 2.300 L2 . atm/m012 and b 0.0430 L/mol Express your answer with the appropriate units The volume of a gas is defined as the space in which the gas molecules can move. When using the ideal gas law we space is equal to the volume of the container However, the gas molecules themselves ta some space in the container. So in the van der Waals equation, the variable V is adjusted to be the volume of the container minus the space taken up by the molecules, V-nb, where b is the volume of a mole of molecules make the assumption that this atm/mol an ke up IA pressure differenceValue Units pressure difference =

Explanation / Answer

Answers , -

Part A ..................H2

Part B..................CO

Explanation-

The inferance can be based on the prediction of compressibility factor ( Z ) of the given gases.

When Z = 1 , the gas behaves as an ideal gas . Since ,Z = PV / nRT

considering the extreme conditions of temperature( T ) and pressure ( P ), the value of Z should depend upon the number of moles present in a volume V. The number of moles , on the other hand , depends on molar mass of the gas. Greater the molar mass lesser should be number of

moles present in a definite volume V.

Part C................. 2.68 atm.

Calculations

under ideal behaviour-

Pressure is calculated using relation PV = nRT

substituting the given values

n = 12 , R = 0.08206 L atm / mol K , T = 293.9 K V = 8.00 L ......we get ,

P = n/V RT

... = 12 /8 [ 0.08206 x 293.9 ]

... = 36.17632 atm.

Again , assuming real gas behaviour of the gas pressure under given conditions are calculated using van - der Waal's equation,

( P + an2 / V2 ) ( V - nb ) = n RT

putting up the given values , we get

{ P + [ 2.30 x (122 ) / 82 ] } { (8 - 12 x 0.0430 ) ] = 12 x 0.08206 x 293.9

where , a = 2.30 L2 atm. / mol2   & b = 0.0430 L / mol

Solving for P we get , P= 33.49539 atm.

Hence , the difference in pressures = ( 36.1762 - 33.49539 ) atm.

..........................................................= 2.6808 atm.

.................................................. or ,= 2.68 atm.

Further the unit of pressure is determined as "atm'' through dimensional analysis.

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