Boyle’s Law
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Transcript of Boyle’s Law
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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.
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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
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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)
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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)
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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
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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)
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Charles’ Law states that for a sample of gas at constant P,
V1 = V2 (P, n const)
T1 T2
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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)
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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!
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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
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What is the conversion between Celsius and Kelvin?
K = oC + 273
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ALWAYS do calculations using
absolute, or KELVIN TEMPERATURE
K = oC + 273
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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.
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data obtained in class . . .
Temp (oC) Pressure (psi)
79 16.9
24 14.7
0 13.6
-196 4.0
-79 9.8
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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
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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
)
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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.
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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?
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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
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Combined Gas LawP1V1 = P2V2 for constant n
T1 T2
Useful for calculations involving changes in T, P, V for a fixed amount of gas.
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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)
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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.
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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