Avogadro’s Hypothesis

Recall: 1.00 mol of any gas @ SATP occupies a volume of 24.8 L;

or 2.00 mol of any gas @ SATP occupies

49.6 L;

or 0.500 mol of any gas @ SATP occupies

12.4 L.

How will a graph of V vs n look @SATP?

Linear, with slope = 24.8 L/mol.

Therefore:

V α n, at fixed T, P;

V1 = V2 (P, T unchanged), or

n1 n2

V = constant (P, T unchanged)

n

Putting Avogadro’s Hypothesis into the mix: gas law eq’n unchanged

Boyle’s P1V1 = P2V2 n, T

Charles’ V1 = V2 n, P

T1 T2

Gay-Lussac’s P1 = P2 n, V

T1 T2

Avogadro’s V1 = V2 P, T

Hypothesis n1 n2

P1V1 = P2V2 = P3V3 = constant value

n1T1 n2T2 n3T3

PV = R ideal gas constant

Tn

What is the value of R?

Use the molar volume of any gas @ SATP to calculate R.

P = 101.3 kPa

T = (25 + 273) = 298 K

V = 24.8 L

n = 1.00 mol

R = PV/(nT) = 8.31 kPa*L*mol-1*K-1

What if P = 760 mmHg? Value of R will be . . .

R = 62.4 mmHg*L*mol-1*K-1

What if P = 1.00 atm? Value of R will be . . .

R = 0.0821 atm*L*mol-1*K-1

Ideal Gas LawRecall

PV = R cross-multiplying gives

Tn

PV = nRT IDEAL GAS LAWFine print:

Valid for any ideal gas, BUT NOT at

high P or low T—because gases will

liquefy.

sample problem #1What volume is occupied by 25.2 g of Xenon

contained at a pressure of 862 kPa at 30oC?

PV = nRT

n = 25.2 g/131.3 g*mol-1 = 0.192 mol

P = 862 kPa

T = 30 + 273 = 303 K

R = 8.314 L*kPa*mol-1K-1

V = nRT/P = 0.561 L

sample problem #2What is the pressure (in atm) of propane,

C3H8, at 20oC, contained in a 1.0 L tank, if the tank holds 465 g of C3H8?

(101.3 kPa = 1.00 atm)n = 465 g/(44 g*mol-1) = 10.6 molT = (20 + 273) = 293 KR = 0.0821 L*atm*mol-1*K-1

V = 1.0 LP = nRT/V = 255 atm

Holy $%^%!!

We calculated the P inside to be

255 atm!

Can this be right?

Shake the cylinder.

What do you hear?

Liquid propane sloshing around.

What does this suggest about the actual P inside?

Less than calculated—C3H8 not all in gas phase.

Aside: A propane explosion looks like this . . .

http://www.youtube.com/watch?v=__1Ym_F94CE

sample problem #3 (gas stoichiometry)

What volume (L) of hydrogen gas, can be expected from the reaction of 12.3 g of Mg with 450 mL of 0.976 mol/L HCl at 25oC at a pressure of 98.4 kPa?

Where do we start?

Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g)

Now what?

Find limiting reagent.

Mg(s) + 2 HCl(aq)

12.3 g 0.450 L

↓/24.3 g*mol-1 * 0.976 mol/L

= 0.506 mol = 0.439 mol

↑limiting reagent

2 HCl(aq) H2(g)

0.439 mol *1/2 0.220 mol expected

Use PV = nRT, or

V = nRT/P to get the volume of H2

= [(0.220 mol)*(8.314 L*kPa*mol-1K-1)*(298 K)]

98.4 kPa

= 5.54 L of H2 expected

HW

PP on p 556—do a few (see previous worked examples if you need help)

PP on p 560 #31- 38