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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