Charles law

7 Key excerpts on "Charles law"

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  eBook - ePub

  Chemistry

  Concepts and Problems, A Self-Teaching Guide

  • Richard Post, Chad Snyder, Clifford C. Houk(Authors)
  • 2020(Publication Date)
  • Jossey-Bass

    (Publisher)

  With the pressure constant, the volume of a gas is directly proportional to the temperature. (That is, if the volume is multiplied by a number, the temperature must be multiplied by that same number. If the volume is divided by some number, the temperature must also be divided by that number.)

  A scientist named Charles first stated this relationship formally. Charles's Law can be stated mathematically as:

  V represents the volume of a gas and T is the absolute temperature of a gas.

  1. In Boyle's Law, at constant temperature, the pressure and volume are (directly, inversely) ________________ proportional.
  2. In Charles's Law, at constant pressure, the volume and absolute temperature are (directly, inversely) ________________ proportional.

  Answer: (a) inversely; (b) directly

  Charles's Law can be rewritten as:

  V1

  is the original volume and

  V2

  is the new volume.

  T1

  is the original temperature and

  T2

  is the new temperature.

  A quantity of gas at 200 K (absolute temperature scale) is heated to 400 K. The original volume is 3 liters. What is the new volume (with no change in pressure)?____________________________________________

  Answer: Charles's Law applies here. Modify the equation so that just V2 is on the left side of the equation by multiplying both sides by T2 .

  Charles's Law is based on absolute temperature. The absolute (Kelvin) temperature scale is comparable to the Celsius scale in size of the degree, but the zero point is different. The zero point for the Kelvin scale is known as absolute zero. The zero point for the Celsius scale is the freezing point of water. The Kelvin scale zero point (0 K) is equivalent to −273°C. The freezing point of water is 0°C, or 273 K. To change from Celsius to Kelvin, just add 273 to the Celsius temperature. To change from Kelvin to Celsius, just subtract 273 from the Kelvin temperature. Fill in the blanks in the following conversions.

  0°C = 273 K 10°C = 283 K 50°C = ___________K _______ = 300 K

  Answer: 323; 27°C

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  eBook - ePub

  From Vehicles to Grid to Electric Vehicles to Green Grid

  Many a Little Makes a Miracle

  • Fuhuo Li, Shigeru Kanemitsu;Jianjie Zhang(Authors)
  • 2019(Publication Date)
  • WSPC

    (Publisher)

  times its initial volume at time 0.

  It is convenient to introduce the new unit of temperature, absolute temperature, in K (Kelvin) introduced by Lord Kelvin (neé Thomson) and denoted by T. Between the two unit systems Celsius and absolute temperature, the relation

  holds. In view of (2.2), the Charles’ law (2.1) may be written as Two special cases of (2.2) are noteworthy:

  which is called the absolute zero and there cannot exist temperature lower than this;

  There is another governing law, the Boyle’s law which says that under constant temperature, pressure and volume (of a gas) are inversely proportional:

  In thermodynamics, such gases that satisfy Boyle’s law are called ideal gases. Examples of ideal gases are Helium He, hydrogen H2 , oxygen O2 , nitrogen N2 etc., which are not easily liquidated.

  Given n mol ideal gas, the Boyle-Charles law

  holds, where R is the gas constant given by

  (2.6) is called the equation of state of the ideal gas.

  Let M and M denote the mass and molecular weight of the gas, then and (2.6) may also be written as

  A more advanced state equation than (2.6) is the van der Waals state equation

  where a > 0 and b > 0 are constants intrinsic to the gas.

  2.2Engine terminology

  We assemble here some of the most important terminologies on the engine.

  •Vc designates the clearance volume, which is the volume of the cylinder when the piston is at T.D.C = Top Dead Center (or T.D.P. = Top Death Point). This is the volume which always remains in the cylinder, i.e. the mixed gas cannot be compressed more than this.

  •L means the stroke,

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  eBook - ePub

  Understanding General Chemistry

  • Atef Korchef(Author)
  • 2022(Publication Date)
  • CRC Press

    (Publisher)

  2 is unknown, we have:

  C = V 1 T 1 = V 2 T 2

  Rearranging gives

  T 2

  =

  V 2

  ×

  T 1

  V 1

  The temperature should be expressed in K, so T1  = 125°C = (125 + 273) K = 398 K.

  Solving gives

  T 2

  =

  1 .6   L × 398   K

  3 .2   L

  =199   K

  It is interesting to note that the volume is reduced by a factor of two and that the temperature decreases by the same factor, two.

  10.5 Avogadro’s Law

  Avogadro’s law states that, at a constant temperature and pressure, the volume of the ideal gas is directly proportional to gas amount (number of moles n), that is n/V = constant. This means that, when the gas amount increases by a certain factor, the gas volume increases by the same factor; conversely, when the gas amount decreases by a certain factor, the volume of the gas decreases by the same factor. At a constant temperature and pressure, when the gas amount increases, the gaseous molecules collide with the walls of their recipient (and with one another) more often. The increase in the number of collisions increases the pressure, which forces the gaseous molecules to occupy a larger volume to increase the wall surfaces and hence to decrease the number of collisions, until the collective impact of the collisions of the molecules with the recipient walls precisely balances the applied pressure.

  Practice 10.3 A balloon contains 1.1 mol of helium (He) occupying 26.2 L. What volume will the gas occupy when 1.21 mol of He was added to the balloon if the pressure and temperature remain constant?

  Answer:

  We are looking for a volume change due to a gas amount change in a balloon at constant temperature and pressure, so we will use Avogadro’s law, that is n/V = A, and A is a constant. Taking V1  = 26.2 L and n1  = 1.1 mol as the initial values, and n2 as the number of moles where the volume V2

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  eBook - ePub

  General Chemistry for Engineers

  • Jeffrey Gaffney, Nancy Marley(Authors)
  • 2017(Publication Date)
  • Elsevier

    (Publisher)

  Fig. 6.10B , the pressure and temperature become directly proportional and pressure becomes equal to zero at 0 K.

  Fig. 6.10 The pressure of a gas sample as a function of temperature in degrees Celsius (A) and Kelvin (B).

  The modern statement of Gay-Lussac’s law is;

  •  At constant volume, the pressure of a fixed mass of any gas is directly proportional to the absolute temperature in degrees Kelvin.

  P = k T

    (9)

  This means that since the ratio of pressure to temperature for any gas at constant pressure is (P /T  = k ), if the pressure or temperature of the gas is changed, the effect on the other variable can be calculated by;

  P 1

  /

  T 1

  =

  P 2

  /

  T 2

    (10)

  Example 6.4: Determining the Pressure of a Gas After a Change in Temperature at Constant Volume and Mass If a gas contained in a steel tank at 21.4°C has a pressure of 5.17 atm. What will the pressure be if it is heated to a temperature of 89.6°C? According to Gay-Lussac’s Law:

  P 1

  = 5.17 atm ,

  T 1

  = 21 .

  4 °

  C = 21.4 + 273.15 = 294.6 K

  T 2

  = 37 .

  5 °

  C = 37.5 + 273.15 = 310.7 K

  So

  P 1

  /

  T 1

  =

  P 2

  /

  T 2

  P 2

  =

  P 1

  T 2

  /

  T 1

  =

  5.17 atm

  310.7 K

  294.6 K

  = 5.45 atm

  6.5 The Ideal Gas Law

  Boyle’s law, Charles’ law, and Gay-Lussac’s law each describe relationships between pairs of the three important variables that determine the behavior of a gas (temperature, pressure, and volume). In order to determine the values of all three variables when more than one is changing, the three gas laws can be combined into a single law. This gives a relationship between pressure, volume, and temperature for a fixed amount of any gas expressed as a single equation called the combined gas law

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  eBook - ePub

  Science and Mathematics for Engineering

  • John Bird(Author)
  • 2019(Publication Date)
  • Routledge

    (Publisher)

  Chapter 32 Ideal gas laws

  Why it is important to understand: Ideal gas laws

  The relationships that exist between pressure, volume and temperature in a gas are given in a set of laws called the gas laws, the most fundamental being those of Boyle, Charles, and the pressure or Gay-Lussac’s law, together with Dalton’s law of partial pressures and the characteristic gas equation. These laws are used for all sorts of practical applications, including for designing pressure vessels, in the form of circular cylinders and spheres, which are used for storing and transporting gases. Another example of this is the pressure in car tyres, which can increase due to a temperature increase, and can decrease due to a temperature decrease. Other examples are large and medium size gas storage cylinders and domestic spray cans, which can explode if they are heated. In the case of domestic spray cans, these can explode dangerously in a domestic situation if they are left on a window sill where the sunshine acting on them causes them to heat up or, if they are thrown on to a fire. In these cases, the consequence can be disastrous, so don’t throw your ‘full’ spray can on to a fire; you may very sadly and deeply regret it! Another example of a gas storage vessel is that used by your ‘local’ gas companies, which supply natural gas (methane) to domestic properties, businesses, etc.

  At the end of this chapter, you should be able to:

  • state and perform calculations involving Boyle’s law
  • understand the term isothermal
  • state and perform calculations involving Charles’ law
  • understand the term isobaric
  • state and perform calculations involving the pressure or Gay-Lussac law
  • state and perform calculations on Dalton’s law of partial pressures
  • state and perform calculations on the characteristic gas equation
  • understand the term STP

  Science and Mathematics for Engineering. 978-0-367-2O475-4, © John Bird. Published by Taylor & Francis. All rights reserved.

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  Compressors

  Selection and Sizing

  • Royce N. Brown(Author)
  • 2011(Publication Date)
  • Gulf Professional Publishing

    (Publisher)

  Many of the common “gases” used in compressors for process plant service are actually vapors. In many cases, the material may change states during a portion of the compression cycle. Water is a good example, since a decrease in temperature at high pressure will cause a portion of the water to condense. This is a common occurrence in the first intercooler of a plant air compressor. Conversely, lowering the pressure in a reservoir of liquid refrigerant at a fixed temperature will cause the vapor quantity to increase.

  Perfect Gas Equation

  Jacques A. C. Charles and Joseph Gay-Lussac, working independently, found that gas pressure varied with the absolute temperature. If the volume was maintained constant, the pressure would vary in proportion to the absolute temperature [4] . Using a proportionality constant R, the relationships can be combined to form the equation of state for a perfect gas, otherwise known as the Perfect Gas Law.

  (2.1)

  where

  P = absolute pressure v = specific volume R = constant of proportionality T = absolute temperature

  If the specific volume v is multiplied by mass m, the volume becomes a total volume V. Therefore, multiplying both sides of Equation 2.1 by m yields

  (2.2)

  In process engineering, moles are used extensively in performing the calculations. A mole is defined as that mass of a substance that is numerically equal to its molecular weight. Avogadro’s Law states that identical volumes of gas at the same temperature and pressure contain equal numbers of molecules for each gas. It can be reasoned that these identical volumes will have a weight proportional to the molecular weight of the gas. If the mass is expressed as

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  Fundamentals of Engineering Thermodynamics

  • V. Babu(Author)
  • 2019(Publication Date)
  • CRC Press

    (Publisher)

  i.e., real gases.

  Figure 5.6:      Region of ideal gas behaviour. Adapted from Thermodynamics: An Engineering Approach by Cengel and Boles.

  5.2.1    Perfect gas equation of state

  The equation of state for an ideal gas may be written as

  P v = R T

  (5.4)

  where T is the temperature. R is the particular gas constant and is equal to

  ℛ ÷ M

  where

  ℛ = 8314   J/kmol .K

  is the Universal Gas Constant and M is the molecular weight of the gas in units of kg/kmol. Equation 5.4 may be written in many different forms depending upon the application under consideration. A few of these forms are presented here for the sake of completeness. Since the specific volume v = 1/ρ, we may write

  P = ρ R T

  or, alternatively as

  P V = m R T

  Figure 5.7:      Compressibility chart for gases

  where m is the mass and V is the volume. If we define the concentration c as (m ÷ M)(1/V), then,

  P = c ℛ T

  Here, c has units of kmol/m3 . The mass density ρ is be related to the particle density n (particles/m3 ) through the relationship ρ = nM ÷ NA . Here, we have used the fact that 1 kmol of any substance contains Avogadro number of molecules (NA = 6.023 × 1026 ). Thus

  P = n

  ℛ

  N A

  T = n

  k B

  T

  where kB is the Boltzmann constant.

  5.2.2    Calorically perfect gas

  In thermodynamics, we need, in addition to the equation of state, an equation relating the internal energy to other measurable properties. The internal energy, strictly speaking, is a function of two thermodynamic properties, namely, temperature and specific volume. In reality, the dependence on specific volume is very weak for gases and hence is usually neglected. Such gases are called thermally perfect and for them u = u (T

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