boyle’s law imagine: hold your finger over the hole at the end of a syringe. now depress the...

Boyle’s Law

Imagine:

Hold your finger over the hole at the end of a syringe.

Now depress the plunger.

How does the pressure of the gas in the syringe change as the volume of gas decreases?

The P increases.

Data from Boyle’s Law Exp’t.

P (kPa) V(mL) 1/V (mL-1)P*V(kPa*mL)

100 100 0.01 10 000

125 80 0.0125 10 000

167 60 0.017 10 000

250 40 0.025 10 000

500 20 0.050 10 000

We can plot two graphs of these data:

P1V1 = P2V2 or

PV = k ( a constant)

(T,n unchanged)

Check this . . .

Boyle's Law Data (Const T, n)

0

100

200

300

400

500

600

0 20 40 60 80 100 120

Volume of Gas (mL)

Pre

ssur

e of

Gas

(kP

a)

We can also make a reciprocal plot:

P α 1/V or

P = k * (1/V) or

P = k/V or

PV = k

(T, n unchanged)

Boyle's Law Data-Reciprocal Plot (Const T, n)

R2 = 1

0100200300400500600

0 0.01 0.02 0.03 0.04 0.05 0.06

1/Volume (mL-1)

Pre

ssur

e (k

Pa)

Charles’ Law

Jacques Charles collected the following data for a sample of gas maintained at constant Pressure:

He obtained the same data when different gases were used.

V (L) T (oC)

0.75 -192

1.0 -165

1.5 -111

2.0 -57

2.5 -3.6

3.0 50.2

3.5 104

Graphing Charles’ Data . . .V α T (P, n unchanged) or

V = (constant)*T or

V = constant or

T

V1 = V2 (P, n const)

T1 T2

Charles Law Data y = 0.0093x + 2.5328

R2 = 1

0

1

2

3

4

-250 -200 -150 -100 -50 0 50 100 150

Temperature (oC)

Volu

me

(L)

Charles’ Law states that for a sample of gas at constant P,

V1 = V2 (P, n const)

T1 T2

Look again at Charles’ Law Data:

Equation of line is: V = 0.0093T + 2.5328

What will be the temperature when the gas occupies zero volume?

T = - 2.5328/0.0093 = -272oC This is the

Absolute zero of temperature.

(Accepted value 0 K = - 273oC)

Charles Law Data y = 0.0093x + 2.5328

R2 = 1

0

1

2

3

4

-250 -200 -150 -100 -50 0 50 100 150

Temperature (oC)

Vo

lum

e (

L)

Temperature ScalesCharles’ Law states V α T (@ constant n, P)

If T ↑ 2X, therefore

V ↑ 2X.

But what is 2X (-10oC) ?

-20oC ??? Colder??? Doesn’t make sense!

What is 2X 0oC?

0oC??? Doesn’t make sense!

So what’s up with the Celsius Temperature scale?

It’s relative to water:

0oC is fp of water; 100oC is bp of water.

But—the Kelvin Scale is

What is the conversion between Celsius and Kelvin?

K = oC + 273

ALWAYS do calculations using

absolute, or KELVIN TEMPERATURE

K = oC + 273

Gay-Lussac’s Law

Henri Gay-Lussac studied the relationship between P and T for a fixed volume of gas.

Refer to class demo using above apparatus.

data obtained in class . . .

Temp (oC) Pressure (psi)

79 16.9

24 14.7

0 13.6

-196 4.0

-79 9.8

Graphing these data gives:Gay-Lussac's Law y = 0.0474x + 13.43

R2 = 0.9985

0

5

10

15

20

-350 -250 -150 -50 50 150

Temperature (oC)

Pres

sure

(psi

)

P α T (V, n unchanged) or

P = (constant)*T or

P = constant or T

P1 = P2 (V, n unchanged)T1 T2

Looking at the graph again . . .

Equation of line isP = 0.0474*T + 13.42

Predict the T at which Pgas = 0?

Gay-Lussac's Law y = 0.0474x + 13.43

R2 = 0.9985

0

5

10

15

20

-350 -250 -150 -50 50 150

Temperature (oC)

Pre

ssu

re (

psi

)

P = 0.0474*T + 13.42, set P = 0 and solve for T

0 = 0.0474*T + 13.42T = -13.42/0.0474 T = -283oC not bad for class data!

Absolute zero (-273oC or 0 K )can be determined from either

V vs T or P vs T.

check this . . . “. . . circulate to all engineers showing the folly of poor design and/or incorrect operating procedures. Apparently the rail car had been steamed out and was still hot inside when it started to rain. The tank had a vent designed to release pressure, not for a vacuum. “

What do you think happened to the rail car?

So far:

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

Put these together to get the

Combined Gas Law:

P1V1 = P2V2 n unchanged

T1 T2

Combined Gas LawP1V1 = P2V2 for constant n

T1 T2

Useful for calculations involving changes in T, P, V for a fixed amount of gas.

example When a 12.0 L sample of H2S(g), originally

at 101kPa and 25oC is subjected to a pressure of 205 kPa at 78oC, what will be the “new” volume of the gas?

Use combined gas law:

P1 = 101 kPa

V1 = 12.0L

T1 = (25 + 273 = 298K)

P2 = 205 kPa

T2 = (78 + 273 = 351K)

P1V1 = P2V2

T1 T2 rearranging gives

V2 = P1V1T2

P2T1

= (101 kPa)(12.0 L)(351 K)

(205 kPa)(298 K)

= 6.96 L is the expected volume.

HW

Text p 511 to 542

P 514 PP – do a few

P 515 RQ #1 – 14 do the ones that challenge you

P 518 LC #13 – 18

P 522 PP – do a few

P 525 PP – do a few

P 542 PP 1 – 10 do a few