Chapter 10 Temperature and Kinetic Theoryroselleachs.sharpschool.net/UserFiles/Servers/Server...Gas Laws, Absolute Temperature, and the Kelvin Temperature Scale Thermal Expansion The
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10.3 Gas Laws, Absolute Temperature, and the Kelvin Temperature Scale
A constant-volume gas thermometer is useful because the temperature is directly proportional to the pressure. If P-T curves are plotted for different gases, they converge at zero pressure.
10.4 Thermal Expansion Most materials expand when heated. For small changes in temperature, the change in length is proportional to the change in temperature.
Two drinking glasses are stuck, one inside the other. How would you get them unstuck?
a) gets larger
b) gets smaller
c) stays the same
d) vanishes
Metals such as brass expand when heated. The thin brass plate in the movie has a circular hole in its center. When the plate is heated, what will happen to the hole?
Question 10.5b Steel Expansion II
10.4 Thermal Expansion Water behaves nonlinearly near its freezing point—it actually expands as it cools. This is why ice floats, and why frozen containers may burst.
10.6 Kinetic Theory, Diatomic Gases, and the Equipartition Theorem
The atoms in a monatomic gas have only translational equilibrium to contribute to the internal energy. A diatomic molecule can also rotate around two distinct axes (x and y).
Boyle’s Law Boyle’s Law states that • The pressure of a gas
is inversely related to its volume when T and n are constant.
• If the pressure increases, volume decreases.
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In Boyle’s Law • The product P x V is constant as long as T and n do
not change. P1V1 = 8.0 atm x 2.0 L = 16 atm L P2V2 = 4.0 atm x 4.0 L = 16 atm L P3V3 = 2.0 atm x 8.0 L = 16 atm L
• Boyle’s Law can be stated as P1V1 = P2V2 (T, n constant)
PV Constant in Boyle’s Law
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Solving for a Gas Law Factor
The equation for Boyle’s Law can be rearranged to solve for any factor.
P1V1 = P2V2 Boyle’s Law To solve for V2 , divide both sides by P2.
P1V1 = P2V2 P2 P2
V1 x P1 = V2 P2
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Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 8.0 L sample of Freon gas initially at 550 mm Hg after its pressure is changed to 2200 mm Hg at constant T?
STEP 1 Set up a data table Conditions 1 Conditions 2 P1 = 550 mm Hg P2 = 2200 mm Hg V1 = 8.0 L V2 =
Calculation with Boyle’s Law
?
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STEP 2 Solve Boyle’s Law for V2. When pressure increases, volume decreases. P1V1 = P2V2
V2 = V1 x P1 P2 V2 = 8.0 L x 550 mm Hg = 2.0 L 2200 mm Hg pressure ratio decreases volume
Calculation with Boyle’s Law (Continued)
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Learning Check For a cylinder containing helium gas indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant): 1) Pressure decreases 2) Pressure increases
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Solution For a cylinder containing helium gas indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant): 1) Pressure decreases B 2) Pressure increases A
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Charles’ Law In Charles’ Law • The Kelvin temperature
of a gas is directly related to the volume.
• P and n are constant.
• When the temperature of a gas increases, its volume increases.
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• For two conditions, Charles’ Law is written V1 = V2
(P and n constant)
T1 T2
• Rearranging Charles’ Law to solve for V2
V2 = V1 x T2
T1
Charles’ Law: V and T
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Learning Check
Solve Charles’ Law expression for T2. V1 = V2 T1 T2
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Solution
V1 = V2 T1 T2 Cross multiply to give V1T2 = V2T1 Isolate T2 by dividing through by V1 V1T2 = V2T1 V1 V1 T2 = T1 x V2 V1
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• The relationship between the four properties (P, V, n, and T) of gases can be written equal to a constant R. PV = R nT
• Rearranging this expression gives the expression called the ideal gas law. PV = nRT
Ideal Gas Law
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The universal gas constant, R • Can be calculated using the molar volume of a gas at STP. • Calculated at STP uses 273 K,1.00 atm, 1 mole of a gas, and a
molar volume of 22.4 L. P V R = PV = (1.00 atm)(22.4 L) nT (1 mole) (273K) n T = 0.0821 L atm mole K
Universal Gas Constant, R
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Another value for the universal gas constant is obtained
using mm Hg for the STP pressure. What is the value
of R when a pressure of 760 mm Hg is placed in the R
value expression?
Learning Check
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What is the value of R when a pressure of 760 mm Hg is placed in the R value expression?
R = PV = (760 mm Hg) (22.4 L) nT (1 mole) (273 K) = 62.4 L mm Hg
mole K
Solution Question 10.9a Ideal Gas Law I
a) cylinder A
b) cylinder B
c) both the same
d) it depends on temperature T
Two identical cylinders at the same
temperature contain the same gas. If
A contains three times as much gas
as B, which cylinder has the higher
pressure?
Question 10.9b Ideal Gas Law II
a) cylinder A
b) cylinder B
c) both the same
d) it depends on the pressure P
Two identical cylinders at the same pressure contain the same gas. If A contains three times as much gas as B, which cylinder has the higher temperature?
Question 10.9c Ideal Gas Law III Two cylinders at the same temperature contain the same gas. If B has twice the volume and half the number of moles as A, how does the pressure in B compare with the pressure in A?