chapter 5 gases -1 gas laws - manu's...

Chapter 5Gases -1Gas Laws

Dr. Sapna Gupta

Properties of Gases

1. They are compressible.

2. They expand to fill the container.

3. Pressure, volume, temperature, and amount are related.

4. They have low density.

5. Form a homogeneous mixture.

Dr. Sapna Gupta/Gases 2

Kinetic Molecular Theory

A theory, developed by physicists, that is based on the assumption that a gas consists of molecules in constant random motion.

Postulates of the Kinetic Theory

1. Gases are composed of molecules whose sizes are negligible.

2. Molecules move randomly in straight lines in all directions and at various speeds.

3. The forces of attraction or repulsion between two molecules (intermolecular forces) in a gas are very weak or negligible, except when the molecules collide.

4. When molecules collide with each other, the collisions are elastic.

5. The average kinetic energy of a molecule is proportional to the absolute temperature.


Pressure

Pressure is force exerted per unit area. P = Force/Area

The SI unit for pressure is the pascal, Pa.

Other Units

atmosphere, atm

mmHg

torr

bar


Measurement of Pressure

A barometer is a device for measuring the pressure of the atmosphere.

A manometer is a device for measuring the pressure of a gas or liquid in a vessel.


Gas Laws

• Gas laws – empirical relationships among gas parameters. • Volume (V)

• Pressure (P)

• Temperature (T)

• Amount of a gas (n)

• By holding two of these physical properties constant, it becomes possible to show a simple relationship between the other two properties.


Boyle’s Law• The volume of a sample of gas is inversely proportional to pressure at

constant temperature. (one increases if the other decreases and vice versa)

PV = constant

P1V1=P2V2

PV

1


When a 1.00-g sample of O2 gas at 0oC is placed in a container at a pressure of 0.50 atm, it occupies a volume of 1.40 L.

When the pressure on the O2 is doubled to 1.0 atm, the volume is reduced to 0.70 L, half the original volume.

Boyle’s Law Contd….

Graphical representation


Solved Problem: A volume of oxygen gas occupies 38.7 mL at 751 mmHg and 21°C. What is the volume if the pressure changes to 359 mmHg while the temperature remains constant?

Vi = 38.7 mL

Pi = 751 mmHg

Ti = 21°C

Vf = ?

Pf = 359 mmHg

Tf = 21°C

f

iif

P

VPV

mmHg)(359

mmHg)mL)(751(38.7f V

= 81.0 mL

(3 significant figures)


Charles’s Law

• The volume of a sample of gas at constant pressure is directly proportional to the absolute temperature (K).

TV

At room Temp At – 72oC


f

f

i

i

constant

T

V

T

V

T

V

Charles’s Law contd….

Graphical Representation

The temperature -273.15°C is called absolute zero. It is the temperature at which the volume of a gas is hypothetically zero.

This is the basis of the absolute temperature scale, the Kelvin scale (K).


Solved Problem:You prepared carbon dioxide by adding HCl(aq) to marble chips, CaCO3. According to your calculations, you should obtain 79.4 mL of CO2 at 0°C and 760 mmHg. How many milliliters of gas would you obtain at 27oC?

Vi = 79.4 mL

Pi = 760 mmHg

Ti = 0°C = 273 K

Vf = ?

Pf = 760 mmHg

Tf = 27°C = 300. K

i

iff

T

VTV

K)(273

mL)K)(79.4(300.f V

= 87.3 mL



Avogadro’s Law

The volume of a gas sample is directly proportional to the number of moles in the sample at constant pressure and temperature.


𝑉

𝑛= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

V a n Vi

ni=Vf

nf

Combined Gas Law

The volume of a sample of gas at constant pressure is inversely proportional to the pressure and directly proportional to the absolute temperature.

The mathematical relationship:

In equation form:

Include the Avogadro’s law and it becomes (1=initial and 2= final)

PiVi

niTi

=PfVf

nfTf

P1V1

n1T1

=P2V2

n2T2

P

TV

f

ff

i

ii

constant

T

VP

T

VP

T

PV


OR

Solved Problem:Divers working from a North Sea drilling platform experience pressure of 5.0 × 101 atm at a depth of 5.0 × 102 m. If a balloon is inflated to a volume of 5.0 L (the volume of the lung) at that depth at a water temperature of 4°C, what would the volume of the balloon be on the surface (1.0 atm pressure) at a temperature of 11°C?

Vi = 5.0 L

Pi = 5.0 × 101 atm

Ti = 4°C = 277 K

Vf = ?

Pf = 1.0 atm

Tf = 11°C = 284 K fi

iiff

PT

VPTV

1

f

(284 K)(5.0 10 atm)(5.0 L)

(277 K)(1.0 atm)V

= 2.6 102 L



Ideal Gas Law

Ideal Gas Law• Combining the historic gas laws yields:

• Adding the proportionality constant, R


R =PV

nT

Standard Temperature and Pressure (STP)

• The reference condition for gases, chosen by convention to be exactly 0°C and 1 atm pressure.

• The ideal gas equation is not exact, but for most gases it is quite accurate near STP (760 torr (1 atm) and 273 K)

• An “ideal gas” is one that “obeys” the ideal gas equation.

• At STP, 1 mol of an ideal gas occupies 22.41 L.

at STP

T = 0°C = 273 K

n = 1 mol

P = 1 atm

V = 22.41 L


Solved Problem:A steel cylinder with a volume of 68.0 L contains O2 at a pressure of 15,900 kPa at 23ºC. What is the volume of this gas at STP?

P2 = 1 atm

T1 = 23 + 273 = 296 KT2 = 273 K

V1 = 68.0 L

L 9850 atm 1K 296

K 273L 68.0atm 157.0

1 1 22

1 2

PVTV

T P

atm 0.157 kPa 101.3

atm 1 kPa 15,900 P1

V2 = ?


P1V1

n1T1

=P2V2

n2T2

Solved Problem:A 50.0-L cylinder of nitrogen, N2, has a pressure of 17.1 atm at 23°C. What is the mass of nitrogen in the cylinder?

V = 50.0 LP = 17.1 atmT = 23oC = 296 K

RT

PVn

K)(296Kmol

atmL0.08206

L)atm)(50.0(17.1

n

mass = 986 g(3 significant figures)

mol

g28.02mol35.20mass

= 35.20 mols


Solved Problem: For an ideal gas, calculate the pressure of the gas if 0.215 mol occupies 338 mL at 32.0ºC.

n = 0.215 mol

V = 338 mL = 0.338 L

T = 32.0 + 273.15 = 305.15 K

P = ?

L×atm

0.215 mol 0.08206 305.15 Kmol×K

0.338 L

P

PV = nRT nRT

PV

= 15.928

= 15.9 atm


Gas Density and Molar Mass

• Using the ideal gas law, it is possible to calculate the moles in 1 L at a given temperature and pressure. The number of moles can then be converted to grams (per liter).

Relation to density

• Get n/V on one side (mol/vol)

• Multiply by molar mass M (g/mol)

• Units of density are g/L here.

To find molar mass, rearrange the above equation.


Solved Problem:What is the density of methane gas (natural gas), CH4, at 125°C and 3.50 atm?

Mm = 16.04 g/molP = 3.50 atmT = 125°C = 398 K

RT

PMd m

K)(398Kmol

atmL0.08206

atm))(3.50mol

g(16.04

d

f igures)tsignifican(3

L

g1.72d


Key Points

• Properties of gases

• The kinetic molecular theory

• Gas pressure

• Units

• Calculation

• Measurement

• The gas laws

• Boyle’s law

• Charles’ law

• Avogadro’s law

• The ideal gas law

• Combined gas law

• Calculating density and molar mass


chapter 5 gases -1 gas laws - manu's...

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