Boyle's Law and Charles' Law

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<p>Area: Chemistry</p><p>Curricular Component: Physical Chemistry I</p><p>Theme: Empirical Properties of Gases</p><p>Summary</p><p>1. Boyle's Law and Charles' Law</p><p>As the gaseous state allows us to have a simpler description in a comparative way, we will</p><p>assume that the system is in equilibrium, to focus on the relationship of mainly four properties: mass,</p><p>pressure, volume and temperature, these properties have not changed during the time since the</p><p>external links of the system do not undergo changes.</p><p>The system for physical chemistry is something that is delimited, a specific part that is being</p><p>observed, studied, in analysis. For a system to have its conditions defined, all the properties that make</p><p>it up must have their values, defined, directly determined by the state of the system. Therefore, the</p><p>state of the system can be described by the values , of some or all of the properties, the more properties</p><p>we describe, with more significant figures, the more accurate we will be. However, generally the</p><p>necessary properties are the four already mentioned: mass (m), pressure (P), volume (V) and</p><p>temperature (t). The State Equation of the System, in turn, is described in a mathematical relationship</p><p>by the four properties, however only three need to be specific (V, P, T) to describe it, as the fourth (m)</p><p>can be obtained from the Equation of State State based on experimental knowledge of the system.</p><p>Robert Boyle in 1662, brought the first quantitative measurements considering the behavior</p><p>between pressure and volume of gases, the results obtained by him reveal that volume (V) is inversely</p><p>proportional to pressure (P) where the product results in a constant (C) . Boyle's law is written as:</p><p>PV = C or P = C/V (eq.1)</p><p>This law only applies to a mass of gas where the temperature is constant.</p><p>The law stated by Charles says that this constant (C) is a function of temperature. Gay-Lussac</p><p>contributed by measuring V while maintaining a fixed mass of gas under a fixed P, discovering that</p><p>V varies linearly with temperature (t) where:</p><p>V = a + bt , (eq. 2)</p><p>where a (the linear coefficient) and b (the angular coefficient) are constants</p><p>We see in the graph to the side that:</p><p>a = V0</p><p>b = (∂V/ ∂t)p</p><p>Note: note that when representing that b is equal to the derivative, the subscript p indicates that the pressure remains</p><p>constant, it does not vary. That's why we use partial derivative instead of common.</p><p>Then we can rewrite eq. 2 like:</p><p>V = V0 + (∂V/∂t)p t (eq. 3)</p><p>From the experience developed by Charles, expressing that for a fixed mass at a fixed pressure,</p><p>the relative increase in volume per degree of temperature was exactly the same for all the gases he</p><p>had used for his measurements. At constant pressure, the increase in volume per degree is given by</p><p>(∂V/ ∂t)p. At 0ºC we have the coefficient of thermal expansion (α0) which is the relative variation in</p><p>volume (ΔV/V) per unit variation in temperature (ΔT), applying the derivative we have:</p><p>Rewriting eq. 3 in terms of α0 we obtain:</p><p>α0 = (1/V0)(∂V/∂t)p (eq. 4)</p><p>Evidenced α0 , 1= α0/α0 , therefore:</p><p>V = V0 α0 (1/ α0 + t ) (eq. 5)</p><p>Everything is in compliance since the volume of the gas is expressed in terms of volume at zero</p><p>degrees and a constant that is the coefficient of thermal expansion (α0) which remains the same</p><p>regardless of the gas and the pressure (p) that is measured, having a limit value for p = 0. Therefore,</p><p>it is suggested that there is a transformation of coordinates, so a new temperature (T) is defined in</p><p>terms of the one used previously, where:</p><p>T= 1/α0 + t (eq. 6)</p><p>This new temperature scale defined by eq. 6, is called the ideal gas scale, which is relevant because</p><p>α0 has the same value for all gases and depends on the originally used temperature scale t.</p><p>Considering that t is in degrees Celcius (ºC), then 1/α0 = 273.15 ºC, it can be seen that the resulting T</p><p>scale is identical to the thermodynamic temperature scale, Kelvin (K), also called absolute</p><p>temperature.</p><p>T= 273,15 + t (eq. 7)</p><p>Combining eq. 5 and eq.6 it is observed that the volume of a gas under a fixed pressure is</p><p>directly proportional to the thermodynamic temperature.</p><p>V = V0 α0 T (eq. 8)</p>