Dr Roger Bennett R.A.Bennett@Reading.ac.uk Rm. 23 Xtn. 8559 Lecture 6 Partial Derivatives • For functions of several variables • z=z(x,y), z is continuous function of independent variables x and y. Here z is called an explicit function. • This function can be represented as a surface in Cartesian coordinates. Partial Derivatives • Imagine the surface intersects a planar surface parallel to the z-x plane which cuts the y-axis at y. • The gradient of the line of intersection is given by the partial derivative of z with respect to x, and is defined as:- z z z ( x x, y ) z ( x, y ) lim x 0 x x y x Partial Derivatives • This is the same expression as for normal differentials but with y considered as a constant. • It effectively gives the gradient in the x direction. A similar expression can be written for the y direction. z z z ( x x, y ) z ( x, y ) lim x x y x0 x Partial Derivatives - example • Let z= x2+3xy +y3 z ( x x, y) ( x x) 3 y( x x) y 2 3 z( x x, y) x 2 xx x 3 y( x x) y 2 2 z z z ( x x, y ) z ( x, y ) lim x x y x0 x 3 Partial Derivatives - example • Ignoring terms second order in the limit x0 z( x x, y) x 2 xx 3 y( x x) y 2 3 z ( x, y) x 3xy y 2 3 z z 2 xx 3 yx lim 2x 3 y x x y x0 x Partial Derivatives • Differential form take the implicit function f(x, y, z) = 0, means x, y,and z are related and only two variables are independent so x = x(y,z) x x dx dy dz z y y z • Similarly we can write y = y (x,z) and y y dy dx dz x z z x Partial Derivatives x x dx dy dz z y y z y y dy dx dz x z z x • Substituting for dy x y x x y dx dx dz y x y z z z z x y z Partial Derivatives x y x x y dx dx dz y z z y z x z x y z • We can choose x and z to be independent variables so choosing dz = 0 is valid, dx can then be non-zero. x y dx dx y z x z x y 1 y z x z 1 x y Reciprocal relation x z y z Partial Derivatives x y x x y dx dx dz y z z y z x z x y z • We can choose x and z to be independent variables so choosing dx = 0 is valid, dz can then be non-zero. x y x 0 y z z x z y x z y 1 Cyclical relation y z x y z x Partial Derivatives – Chain rule • Supposing variables x, y, z, are not independent (so any two variables define the third) we can define the function = (x, y) and re-arrange to x = x(, y). x x dx d dy y y • Dividing by dz and hold constant. x x y z y z Chain Rule Higher Partial Derivatives • Repeated application allows the definition of higher derivatives. z z 2 x x x y 2 z z yx y x y 2 z z xy x y x 2 Higher Partial Derivatives • The order of differentiation does not matter as long as the derivatives are continuous. z z z z yx y x y x y x xy 2 2 Pfaffian forms • The differential equation df Xdx Ydy Zdz 2 • In general the integral df depends on the path of integration. In this case this is 1 called an inexact differential. • But the integral is path independent if it can be expressed as a single valued function f(x,y,z). • df is then an exact differential. Pfaffian forms – exact differentials • If we write f f X Y x y , z y x , z f Z z x , y • It follows from the double differentials that X y Y x , z x y , z Z Y z y , x y z , x Z X x z , y z x , y • These are necessary and sufficient for the Pfaffian form to be an exact differential. Dr Roger Bennett R.A.Bennett@Reading.ac.uk Rm. 23 Xtn. 8559 Lecture 7 Processes – functions of state • The internal energy U of a system is a function of state. For a fluid (ideal gas) we could write U=U(P,T) or U=U(V,T). Processes – functions of state • The internal energy U of a system is a function of state. For a fluid (ideal gas) we could write U=U(P,T) or U=U(V,T). • It is straightforward to imagine that in changing from state 1 to state 2 we can define U1 and U2 such that the difference in energies U=U2-U1 is the change of energy of the system. Processes – functions of state • The internal energy U of a system is a function of state. For a fluid (ideal gas) we could write U=U(P,T) or U=U(V,T). • It is straightforward to imagine that in changing from state 1 to state 2 we can define U1 and U2 such that the difference in energies U=U2-U1 is the change of energy of the system. • We cannot make the same argument for Work or Heat as you cannot define a quantity of either to be associated with a state. They are only defined during changes. Processes – functions of state • The internal energy U of a system is a function of state. For a fluid (ideal gas) we could write U=U(P,T) or U=U(V,T). • It is straightforward to imagine that in changing from state 1 to state 2 we can define U1 and U2 such that the difference in energies U=U2-U1 is the change of energy of the system. • We cannot make the same argument for Work or Heat as you cannot define a quantity of either to be associated with a state. They are only defined during changes. • Work and heat flow are different forms of energy transfer. Heat and Work • The physical distinction between these two modes of energy transfer is: – Work deals with macroscopically observable degrees of freedom. – Heat is energy transfer on the microscopic scale via internal degrees of freedom of the system. Heat and Work • The physical distinction between these two modes of energy transfer is: – Work deals with macroscopically observable degrees of freedom. – Heat is energy transfer on the microscopic scale via internal degrees of freedom of the system. • First Law of Thermodynamics dU = đW + đQ is always true as it is just conservation of energy – you just have to account for all possible heat and work processes. Work • First Law of Thermodynamics dU = đW + đQ is always true as it is just conservation of energy – you just have to account for all possible heat and work processes. – Stretching a wire (spring) đW = Fdx – Surface tension – bubble đW = dA – Charging a capacitor đW = EdZ where E is the emf and Z stored charge. Analysis of processes • Quasistatic work – đW = -PdV P a b b (Wab ) quasistatic PdV a V Analysis of processes P a • Isothermal Work (ideal gas) b b b V dV (Wab ) Isothermal ,quasistatic PdV nkT a a V Vb (Wab ) isothermal,quasistatic nkT ln Va Heat Transfer • Constant pressure heating – đQ = dU - đW – đQ = dU + PdV P a U U dU dT dV T V V T U U – đQ dT P dV T V V T U U V Cp P T V V T T P b • (đQ/T)P = CP specific heat capacity at constant pressure. V Heat Transfer Ideal Gas Only • Cv = (đQ/T)v = (dU/T)v • Cp = (đQ/T)p U U V Cp P T V V T T P • For ideal gas PV = (-1)U = NRT NR V U 0 T P P V T C p CV R C p CV R Heat Transfer Ideal Gas Only • Cv = (đQ/T)v = (dU/T)v = 3/2 R • Cp = (đQ/T)p C p CV R Cp 5 R 2