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10 Thermal Physics
10.4 Expansion of Gasses
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9. A circular plate made of steel has a radius of 0.1 m at a temperature of 20 °C. Calculate its radius at a temperature of 190 °C. (Coefficient of linear expansion of steel is 1.1e-5 ⁄ °C) A. 0.1004 m
B. 0.1006 m C. 0.1002 m D. 0.1 m E. 0.1007 m
10. By how much should the temperature of an aluminum sample change if its volume is to change by 3.25 % of its volume? (Coefficient of linear expansion of aluminum is 2.4e-5 ⁄ °C)
A. 496.528 °C B. 451.389 °C C. 361.111 °C D. 270.833 °C E. 315.972 °C
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The temperature must be in °K when using this equation. °C and °F can’t be used. There are two special cases of this law called Boyle’s law and Charles Law.
Boyle’s law states that if temperature is kept constant, then the volume of a gas is inversely proportional to its pressure. That is, the product of volume and pressure of a gas remains constant.
P1 V1 = P2 V2
Charles’ law states that if pressure of a gas is kept constant, then the volume and the temperature of the gas are directly proportional. That is, the ratio between volume and temperature is a constant.
V1 ⁄ T1 = V2 ⁄ T2
The freezing temperature of water (0 °C) and the atmospheric pressure at sea level (101300 Pa) are commonly referred as standard temperature and standard pressure respectively. The abbreviation STP stands for standard temperature and pressure.
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194
Example: A gas in a balloon has a volume of 20 cm 3 when the temperature is 25 °C and the pressure is 50 kPa. Calculate its volume at STP.
Solution: V1 = 20 cm 3; T1 = 25 °C (T ⁄ °K = T ⁄ °C + 273); P1 = 50, 000 Pa; T2 = 273 °K; P2 = 101, 300 Pa; V2 = ?
T1 ⁄ °K = T1 ⁄ °C + 273 = 25 + 273 = 298 T1 = 298 °K
V1 P1 ⁄ T1 = V2 P2 ⁄ T2
V2 = V1 P1 T2 ⁄ (T1 P2) = 20 * 50, 000 * 273 ⁄ (298 * 101, 300) cm 3 = 9 cm 3 10.4.3 Brief Review of Chemistry
The mass of atoms is measured by a unit of mass called atomic mass unit, abbreviated as u. Atomic mass unit is defined to be (1 ⁄ 12)th of the mass of a carbon atom. The mass of the atom of an element measured in atomic mass unit is called atomic mass (M) of the element. For example the mass of a carbon atom (MC) is 12 u and the mass of an oxygen atom (MO) is 16 u. Molecular mass of a compound is defined to be the sum of the atomic masses of the atoms in the chemical formula of the compound. For example the molecular mass of carbon di-oxide whose chemical formula is CO2 is the sum of the atomic mass of a carbon atom and twice the atomic mass of oxygen: MCO2 = MC +2MO = (12 + 2 * 16) u = 44 u. The gram molecular weight (Mg) of a compound is defined to be its molecular mass expressed in grams (That is Mg = (M ⁄ u) g). For example the gram molecular weight of carbon di-oxide is 44 g because its molecular mass is 44 u. One gram molecular weight of any substance contains 6.02e23 molecules. The number 6.02e23 is called Avogadro’s number and denoted as NA.
NA = 6.02e23
One gram molecular weight of a substance is also called one mole of the substance. Thus the mass of one mole of a substance is equal to gram molecular weight of the substance and there are Avogadro number of molecules in one mole of a substance. The number of moles, n, in a sample of mass m (in grams) may be obtained by dividing the mass of the sample in grams by the molecular weight of the sample, Mg.
n = m ⁄ Mg
Also, the number of moles in a substance can be obtained as a ratio between the number of molecules in the sample, N, and Avogadro number.
n = N ⁄ NA
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Example: A sample of water has a mass of 36 g. The chemical formula of water is H2O. Atomic masses of hydrogen and oxygen are 1 u and 16 u respectively.
a) How many moles are there in the sample?
Solution: m = 36 g; Mg = (2*1 + 16) g = 18 g; n = ? n = m ⁄ Mg = 36 ⁄ 18 = 2 b) How many water molecules are there in the sample?
Solution: N = ?
n = N ⁄ NA
N = nNA = 2 * 6.02e23 = 12.04e23 10.4.4 The Ideal Gas Equation
The combined gas law was stated for a fixed amount of gas. For a situation where the amount of gas also might change, the proportionality between volume and number of moles should also be included. The combined gas law should be restated to say the volume of a gas is directly proportional to temperature and number of moles, and inversely proportional to pressure. Mathematically, this means the ratio between the product of volume and pressure to the product of number of moles and temperature is a constant (PV ⁄ nT = constant). This constant is a universal constant called universal gas constant, denoted by R.
PV = nRT
This is the equation known as the ideal gas equation. The value of R is 8.3 J ⁄ °K ⁄ mole.
R = 8.3 J ⁄ °K ⁄ mole
The ideal gas equation is applicable to real gasses only approximately. It applies exactly only for the ideal case where the interaction energy between the molecules can be neglected completely and the energy of the molecules can be assumed to be purely kinetic energy. In using this equation, SI units (That is m
3 for volume, Pa for pressure and °k for temperature) should be used. But Liter (Abbreviated as L) is a very common unit of volume. If Liter is used for volume, then kilo Pascal (kPa) should be used for pressure. The unit for temperature must be in °K. °C and °F cannot be used.
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Example: Calculate the volume of one mole of any gas at STP in liters.
Solution: n = 1; T = 273 °K; P = 101 Pa; V = ?
PV = nRT
V = nRT ⁄ P = 1 * 8.3 * 273 ⁄ 101 L = 22.4 L Example: 40 L of carbon di-oxide gas is at STP.
a) Calculate the number of moles in this sample.
Solution: V = 40 L; T = 273 °K; P = 101 kPa; n = ? PV = nRT
n = PV ⁄ (RT) = 101 * 40 ⁄ (8.3 * 273) = 1.8 b) How many grams are there in this sample
Solution: Mg = (12 + 2*16) g = 44 g; m = ? n = m ⁄ Mg
m = nM = 1.8 * 44 g = 79.2 g
The ideal gas equation can also be expressed in terms of the number of molecules instead of the number of moles. Replacing n in the ideal equation by N ⁄ NA, The equation PV = NRT ⁄ NA. The ratio R ⁄ NA is a constant called Boltzmann’s constant and is denoted by κβ.
PV = Nκβ T The value of Boltzmann’s constant is 1.38e-23 J ⁄ °K ⁄ mole.
κβ = 1.38e-23 J ⁄ °K ⁄ mole
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Example: A gas has a volume of 0.0002 m 3 at a temperature of 30 °C and a pressure of 20 kPa. How many molecules are there in the sample?
Solution: V = 0.0002 m 3; T = (30 + 273) °K = 303 °K; P = 20 kPa = 20, 000 Pa; N = ? PV = Nκβ T
N = PV ⁄ κβ T = 20, 000 * 0.0002 ⁄ (1.38e-23 * 303) = 1e+21 10.4.5 Energy of Ideal Gas Molecules
An ideal gas is a gas where the interaction energy between the molecules is neglected and the energy is purely kinetic energy. This energy is directly proportional to temperature (in °K) and depends on the number of degrees of freedom of the molecules comprising the gas. Number of degrees of freedom is equal to the number of different ways by which the particles can acquire energy. The random motion of ideal gas molecules can be decomposed into motion along the x-axis, y-axis and z-axis. That is the number of degrees of freedom of the particles is three. The energy per particle per degree of freedom is equal to κβ T ⁄ 2. Therefore, since a particle of an ideal gas has three degrees of freedom, the energy of one particle of an ideal gas is three times the energy per degree of freedom.
E = 3κβ T ⁄ 2
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E is the energy of one molecule of an ideal gas at a temperature T (in °K). The total energy (ET) of the sample is obtained by multiplying the energy of one particle by the number of molecules (N) in the sample.
ET = 3Nκβ T ⁄ 2
The total energy also can be expressed in terms of the number of moles in the sample. Replacing κβ
by R ⁄ NA and noting that the ratio N ⁄ NA is equal to the number of moles n, the following alternative expression for the total energy can be obtained.
ET = 3nRT ⁄ 2
Example: 10 g of hydrogen gas sample (H2) is at a temperature of 20 °C. Atomic mass of hydrogen atom is 1 u
a) Calculate the energy of one molecule.
Solution: T = (20 + 273) °K = 293 °K; E = ?
E = 3κβ T ⁄ 2 = 3 * 1.38e-23 * 293 ⁄ 2 = 6.1e-21 J b. Calculate the total energy of the sample.
Solution: m = 10 g; Mg = 2 * 1 g = 2 g; ET = ?
n = m ⁄ Mg = 10 ⁄ 2 = 5
ET = 3nRT ⁄ 2 = 3 * 5 * 8.3 * 293 ⁄ 2 = 6.1e3 J 10.4.6 RMS speed of Ideal gas molecules
The molecules of a gas are moving randomly with all kinds of speeds. The root mean square speed abbreviated as RMS speed is defined to be the square root of the average of the squares of the speeds of all the particles. The energy of one particle of an ideal gas may also be obtained as the kinetic energy of a particle whose speed is the RMS speed. Thus, if the mass of one molecule is denoted as mo, the equation mo vRMS 2 ⁄ 2 = 3κβ T ⁄ 2 holds; and it follows that
vRMS = √ (3κβ T ⁄ mo)
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m0 (in grams) can be obtained by dividing the gram molecular weight by Avogadro number, since there are Avogadro number of particles in one gram molecular weight. It needs to be in kg though. mo in kg may be given as follows.
mo = Mkg ⁄ (1000NA)
Where Mkg is molecular mass expressed in kg (That is Mkg = (M ⁄ u) kg). Replacing κβ by R ⁄ NA, the following alternative expression for the RMS speed can be obtained.
vRMS = √ (3000RT ⁄ Mkg)
Example: Calculate the RMS speed of a sample of hydrogen gas (H2) at a temperature of 100 °K.
Solution: T = 100 °K; Mkg = 2 kg; vRMS = ?
vRMS = √ (3000RT ⁄ Mkg) = √ (3000 * 8.3 * 100 ⁄ 2) m ⁄ s = 1115.8 m ⁄ s