Ideal-Gas Equation

1

Ideal-Gas Equation

V 1/P (Boyle’s law)V T (Charles’s law)V n (Avogadro’s law)

• So far we’ve seen that

• Combining these, we get

V nTP

2

Ideal-Gas Equation

The relationship

then becomes

nTP

V

nTP

V = R

or

PV = nRT

3

Ideal-Gas Equation

• P = pressure

• V = volume

• n = amount

• T = temperature

• R = gas constant

PV = nRT

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Gas Laws

• The value of the gas constant (R) depends on the units of P, V, n, and T.– T must always be in Kelvin– n is usually in moles

• If P (atm) and V (L), – then R = 0.08206 atm.L

mol.K• If P (torr) and V (L),

– then R = 62.36 L.torr

mol.K

I will give you these on the test.

5

Gas Laws

• The ideal gas law is used to describe the behavior of an ideal gas.

• Ideal gas: hypothetical gas that obeys kinetic molecular theory and the ideal gas law

6

Gas Laws

• The ideal gas law is used in calculations for a specific sample of gas that has a constant T, P, V, and n.

– i.e. no changes are being made to the sample of gas

• If you know 3 of the 4 variables, you can calculate the

other using the ideal gas law.

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Gas Laws

Calculate the volume of 1.00 mol of an ideal gas at 1.00 atm and 0.00oC.

Given: P = 1.00 atm

n = 1.00 mol

T = 0.00oC

= 273 K

Find: V

PV = nRT

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Gas LawsPV = nRT

Solve for V V = nRT

P

Use R = 0.08206 atm.L/molK

V = 22.4 L

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Gas Laws

• The temperature and pressure used in the previous problem are commonly used to report the properties of gases.

• Standard Temperature and Pressure (STP):

0oC and 1 atm

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Gas Laws

• Molar volume: – the volume one mol of a gas occupies

(L/mole)

• At STP, one mole of an ideal gas has a

molar volume of 22.4 L:

22.4 L1 mol

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Think

How many molecules are in 22.4 L of an ideal gas at STP?

Hint: How many moles are there under these conditions?

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Gas Laws

• The ideal gas law applies only to ideal gases.– Does not always accurately describe real gases

• The molar volumes for many real gases at STP differ slightly from 22.4 L/mol.

• In most cases, the differences between ideal gas behavior and real gas behavior is so small that we can ignore it.

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Gas Laws

A weather balloon contains 4.75 moles of He gas. What volume does the gas occupy at an altitude of 4300 m if the temperature is 0.oC and the pressure is 0.595 atm?

Given: P = 0.595 atm

T = 0.oC = 273K

n = 4.75 mol

Find: V

PV = nRT

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Gas Laws

PV = nRT

solve for V V = nRT

P

V=

4.75 mol 0.08206 atm L

mol K0.595 atm

273 K= 179 L

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Gas LawsA used aerosol can contains 0.0173 mol of gas and has a volume of 425 mL. Calculate the pressure in the can if it is accidentally heated to 395oC. (Warning: Don’t do this!!)

Given: V = 425 mL

n = 0.0173 mol

T = 395oC + 273 = 668K

Find: P

PV = nRT

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PV = nRT

solve for P P = nRTV

P = 0.0173 mol 0.08206 atm L

mol K

668 K

425 mL 1 L/1000 mL= 2.23 atm

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Gas Laws

A tire with an interior volume of 3.50 L contains 0.357 mol of air at a pressure of 2.49 atm. What is the temperature of the air in the tire in K? In oC?

Given: P = 2.49 atm

V = 3.50 L

n = 0.357 mol

Find: T (K)

PV = nRT

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Gas Laws

PV = nRT solve for T

T =PV

nR=

2.49 atm 3.50 L

0.357 mol 0.08206 atm L

mol K

= 297 K

= 297 K -273 = 24C

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Gas Laws

• The ideal gas law was useful in determining the properties of a specific sample of gas at constant T, P, V, and n.

• We often need to know how a change in one (or more) properties impacts the other properties for a sample of a gas.

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Combined Gas Laws

• Fixed number of moles of gas (n= constant) and R is a constant:

• PV/T = nR = constant

P1V1 = P2V2

T1 T2

where subscripts mean initial (1) and final (2) conditions of P, V, T

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Gas Laws

A helium-filled balloon occupies 6.00 L at 19.5oC and 0.989 atm. What volume will the balloon occupy on top of Pike’s Peak if the pressure is 0.605 atm and the temperature is constant?

Given: V1 = 6.00L

P1 = 0.989 atm

T1 = T2 = 19.5oC

P2 = 0.605 atm

Find: V2

Since T is constant,

P1V1 = P2V2

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Gas Laws

P1 V1 = P2 V2

(0.989 atm) x (6.00 L) = (0.605 atm) x V2

V2 = 9.81 L

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Gas Laws

Suppose a used aerosol can contains a gas at 0.989 atm at 23oC. If this can is heated to 425oC, what is the pressure inside the can?

Given: P1 = 0.989 atm

T1 = 23oC + 273 = 296K

T2 = 425 + 273 = 698K

V1 = V2

Find: P2

P1V1 = P2V2

T1 T2

= 2.33 atm

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• We know thatmoles molecular mass (g/mol) = mass

Densities of Gases

• So multiplying both sides (of PV = nRT when rearranged to n/V=P/RT) by the molecular mass (M) gives

n M = m

PMRT

mV

=

25

Densities of Gases

• Mass volume = density

• So,

Note: One only needs to know the molecular mass (M), the pressure, and the temperature to calculate the density of a gas.

PMRT

mV

=d =

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Molecular Mass

We can manipulate the density equation to enable us to find the molecular mass of a gas:

Becomes

PMRT

d =

dRTPM =

27

Examples

What is the density of helium gas at 1.00 atm and 25oC?

Given: P = 1.00 atm

T = 25 + 273 = 298K

He gas: 4.00 g/mol

Find: d

d = P M

RT

=1.00 atm 4.00 g mol K

0.08206 atm L 298 K mol= 0.164 g/L

Note: Density of gases units are: g/L

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Gas Laws--More Applications

Example: What is the average molar mass of dry air if it has a density of 1.17 g/L at 21oC and 740.0 torr?

Given: P = 740.0 torr

T = 21 + 273 = 294K

d = 1.17 g/L

Find: M (molar mass)

d = P M

RT

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Solution:

M =

d = P MRT

M = dRTP

1.17 g

740.0 torr

294K 62.36 L torr

mol KL

= 29.0 g/mol

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• Understanding the properties of gases is important because gases are often the reactants or products in a chemical reaction.

– Often need to calculate the volume of gas produced or consumed during a reaction

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Gas Data BPB, VB,

TB

Gas Data APA, VA,

TA

Moles A

Moles B

Molar ratio

grams A

grams B

Molar mass

Molar mass

Moles A

Moles B

Molar ratio

grams A

grams B

Molar mass

Molar mass

PV = nRT

PV = nRT

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The air bag in a car is inflated by nitrogen gas formed by the decomposition of NaN3:

2 NaN3(s) 2 Na (s) + 3 N2 (g)

If an inflated air bag has a volume of 36 L and is to be filled with N2 gas at a pressure of 1.15 atm at a temperature of 26C, how many grams of NaN3 must be decomposed?

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Gas Data

N2

Moles A

Moles B

Molar ratio

grams B

Molar mass

Moles N2

Moles NaN3

Molar ratio

grams NaN3

Molar mass

PV = nRT

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36L N2, 1.15 atm

26oC

Moles A

Moles B

Molar ratio

grams B

Molar mass

Moles N2

Moles NaN3

Molar ratio

grams NaN3

Molar mass

PV = nRT

Given

Find

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n = PV = (1.15 atm) ( 36 L) mol.K

RT (299 K) 0.08206 atm.L

n = 1.7 mol N2

g NaN3 = 1.7 mol N2 x 2 mol NaN3 x 65.0 g NaN3

3 mol N2 1 mol NaN3

g NaN3 = 74 g NaN3

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Example: How many mL of oxygen gas can be collected at STP when 1.0 g of KClO3 decomposes:

2 KClO3 (s) 2 KCl (s) + 3 O2 (g)

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mL O2 at STP

(1.0atm, 0oC)

Moles A

Moles B

Molar ratio

grams B

Molar mass

Moles O2

Moles KClO3

Molar ratio

gramsKClO3

Molar mass

PV = nRT

Find

Given

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Mol O2 = 1.0 g KClO3 x 1 mole KClO3 x 3 mol O2

122.5 g 2 mol KClO3

mol O2 = 0.012 mol O2

(0.012 mol) 0.0821 atm.L (273K) x 103 mL

1.0 atm mol.K 1 L

V = 270 mL

V=nRT

P=

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Ideal-Gas Equation