Gases Chapter 9

What parameters do we use to describe gases?

pressure: force/unit area 1 atm = 101 kPa;

volume: liters (L)

Temperature: K

Amount of gas: moles

A gas consists of small particles that move rapidly in straight linesuntil they collide;they have enough kinetic energy to overcome any attractive forces;gas molecules are very far apart;gas molecules have very small volumes compared to the volumes ofthe containers they occupy;have kinetic energies that increase with an increase in temperature;collisions of the gas cause pressure (force /unit area)

What is meant by % volume? In principle, the actual molecular volume of a gas is so small in comparison to the volume it will occupy that we treat gases at mathematical points.

What is air pressure due to?

A column of air 1 m2 has a mass

of 10,300 kg, producing a

pressure of 101 kPa due to

gravity (14.7 pounds/in2)

1 atm = 76 cm Hg; 101 kPa

Atmospheric pressureis the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth;Pressure is:about 1 atmosphere at sea level; depends on the altitude and the weather;is lower at high altitudes where the density of air is less;

How do we measure pressure?

A barometermeasures the pressure exerted by the gases in the atmosphereindicates atmospheric pressure as the height in mm of the mercury column

A water barometer would be 13.6 times taller than a mercury barometer because the density of Hg is 13.6 times as dense as water

• What is the relationship between the variables used to describe gases?

Note that the initial product between pressure and volume is 4L*1atm = 4 L atm

In the final diagram: 2L*2atm = 4 LatmDecreasing the volume to ½ results in a doubling of the pressure

We find that the product of volume and pressure is equal to a constant

PV = a constant

Initial Final

Boyle’s Law is concerned with the relationship of pressure and volume using a fixed amount of gas ( a fixed number of mols of gas)

P*V = constant at constant temperature

Temperature and Volume

What happens to the number of collisions a molecule makes against the wall if we heat a gas?

the number of collisions increase causing an increase in the pressure inside

What will happen to the volume of the gas if the pressure outside the piston does not change and the pressure inside increases?

Suppose we had a frictionless piston and a gas enclosed within the piston. What could we say about the pressure inside and outside of the piston if the piston was notionless?

pressure P (inside) = P (outside)

temperature (inside) = temperature (outside)

Supppose we now heat the gas inside the piston

V/T(K) = constant

Charles’ Law is concerned with the relationship of temperature and volume when dealing with a constant amount of gas (mols)V ∝ T when T is expressed in K. The K temperature scale is derived from the behavior of gasesif V ∝ T then V = kT where k is a constant at constant pressure

If we allow the pressure to return to it original value, what will happen to the volume?

n is directly proportional to pressure

Pressure and the amount of gas , nSuppose now that we double the number of molecules in the same volume and at the same temperature. What will that do to the number of collisions with the walls? the pressure? What will it do to the volume if we allow the pressure to return to its orginal value?

Avogadro’s Law is concerned with the relationship between the number of molecules or mols (n) and the volume of a gas under conditions of constant pressure and temperature

mols is proportional to volume if T and P is kept constant

Ideal gas law: PV = nRT where R is a constant

R = 0.0821 L.atm/K.mol

Note that at constant n and T, PV = constant Boyle’s Law

Note that at constant P and T, V/n = constant Avogadro’s Law

Note that at constant P and n, V/T = constant Charles’s Law

Summary

Standard conditions of pressure and temperature

T = 0 °C (273 K)

Pressure: 1 atm

What volume does a mol of any ideal gas occupy at STP?

PV = nRTV = 1mol(0.0821 L*atm/K*mol)(273 K)/(1 atm)V = 22.4 L

This means that equal volumes of gases under identical conditions of temperature and pressure contain equal number of molecules

What is the difference between an ideal gas and a real gas?

The ideal gas equation was generated from the kinetic theory of gases making the following assumptions or approximations:

1. The molecules could be treated as points (ie molecular volume = 0)

2. There are no attractive interactions between molecules.

3. Gas particles move around at random

4. Collision of gas molecules with the wall are totally elastic

5. The kinetic energy of the gas particle is ∝ to temperature (K)

In general, the ideal gas law works best at low pressures and high temperatures

Real Gases: van der Waal’s equation

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

an2/V2 corrects for intermolecular attractions

nb corrects for the real volume of molecules

Dalton’s Law of partial pressures:

Total atmospheric pressure = 1 atm;

How much of the pressure is contributed by N2?

Pressure is a consequence of molecules colliding with each other and the walls of the container *062

For air if

If PTV = nTRT and nT = (nO2 + n N2 + ...)

at constant T, PTV = (no2 + n N2 + ...)RT

Since the actual volume of the molecules is small in comparison to the volume occupied by the gas, all molecule occupy the same volume V.

The contribution to the total pressure is dependent on the number of collision of each gas with the wall and this is dependent on the number of molecule of each gas. Hence:

P = (PN2 + PO2 + ...)

PO2V = nO2RT ; PN2V = nN2RT ...

Temperature: a measure of the average kinetic energy of molecules

Distribution of molecular speeds as a function of temperature

What are some of the consequences associated with the fact that molecules at the same temperature have different speeds?

The size of the pinhole needs to be small for effusion

*07

Diffusion: mixing of gases at Effusion: escape the same pressure through a small opening

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Two molecules of different mass at the same temperature effusing through an opening

If M > m, then : 1/2Mv2 = 1/2mV2 V > v

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Two molecules of different mass at the same temperature effusing through an opening

From the kinetic theory of gases

speed of a molecule v = (3RT/M)1/2

For two gases at the same temperature

1/2mava2 = 1/2mbvb

2 va = average speed of molecule a

vb = average speed of molecule b

ma /mb = vb2 /va

2 or vb /va = (ma /mb )1/2

The rate at which molecule a hits the pinhole ∝ v if the comparisons are made at the same concentration and temperature. If you wait long enough ...

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Two molecules of different mass at the same temperature effusing through an opening at equilibrium

Solving some problems involving gases

1. A sample of gas at 25 °C and 2 atm pressure in a 5 L vessel was found to have a mass of 18 g. What is its molecular weight?

PV = n RT

2 atm*5 L = n*0.0821 (Latm/K mol)*298 K

n = 10/(0.0821*298) mol; 0.4087

n = wt/ mw; 0.4087 = 18g/mw

mw = 44 g/mol

Suppose the gas at the right exerted a pressure of 15 cm as shown.

Would the pressure of the gas be greater or less than 1 atm?

How many atm of pressure is the gas exerting?

15.2 cm

1 atm = 76 cm

76 - 15.2 = 60.8

Hg

Suppose the gas at the right exerted a pressure of 15 cm as shown.

Would the pressure of the gas be greater or less than 1 atm?

How many atm of pressure is the gas exerting?

15.2 cm

1 atm = 76 cm

76 - 15.2 = 60.8

Suppose we have a sample of equal amounts of H2 and D2 in a vessel and a small opening is introduced. What will be the initial rates of effusion?

vH2/vD2 = (mD2/mH2)0.5 = (4/2)0.5 = 1.42

Will the relative rate change with time?

What is the density of natural gas (CH4) at STP?

PV = nRT

density is g/mL or g/L

We know the molar volume of any gas is 22.4 L at STP

How many g of methane in a mole?

16g/22.4 L = 0.714 g/l or 7.14*10-4 g/mL or

PV =(wt/mw)*RT; mw*P/RT = (wt/V)

The surface temperature of Venus is about 1050 K and the pressure is about 75 Earth atmospheres. Assuming these conditions represent a “Venusian STP, what is the standard molar volume of a gas on Venus?

PV = nRT

75 atmV =1mol*0.0821(Latm/K mol)*1050 K; V = 1.15 L

Natural gas is a mixture of a number of substances including methane (mol fraction, 0.94); ethane (mol fraction, 0.04); propane (mol fraction, 0.015). If the total pressure of the gases is 1.5 atm, calculate the actual pressure contributed by each of the gases described.

mol fraction = mol A/(mol A + mol B + ....)

PT = 1.5 = P CH4+ PC2H6 + ...

Px V = nxRT nCH4/nC2H6 = PCH4/PC2H6 = 0.94/.04

CH4 = 0.94*1.5

C2H6 = 0.04*1.5

C3H8 = 0.015*1.5