First Law CML 100, IIT Delhi SS. The total energy of the system. Contribution from translation + rotation + vibrations.

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1 Internal Energy he total energy of the system. Contribution from translation + rotation + vibrations. Equipartition theorem for the translation and rotational degrees of freedom. 1/ k B Work Path function, not a state function ðw = p ext d Notation When system does work to the surroundings: negative (expansion) When surroundings does work on the system: positive (compression) wo paths - compression 0 Work done in first path: w = 1 p 1 d 1 p ext d = p 1 ( ) 0 Work done in second path w = p ext d 1 p d p ( ) = Connect the two paths w = p 1 ( ) + p ( ) 0 dw 0 olume olume Heat Another path function olume he First Law du = dq + dw = 0 ΔU = q + w he internal energy of an isolated system is constant. U is a state function. U Does not depend on path. 1

2 We need two variables (other than the number of moles) to define a state function U = U(, ) U is an extensive quantity. However, U/n is intensive Since U = U(, ), we can write du = ( U ) d + ( U ) d In general du = dq + dw exp + dw extra he extra work could be electrical, chemical... Lets keep the volume constant i.e. dw exp = 0 and ensure no additional work i.e. dw extra = 0 hen, du = dq or du = dq We are measuring a change in the internal energy by supplying heat to the system Also, we can write, du = ( U ) d at constant. Define C = ( U ) So, dq = C d Isothermal Expansions a) Free expansion, against vacuum: p ext = 0 dw = p ext d = 0 b) Irreversibe expansion,, dw = p ext d = p d and w = p d = p d = p ( )

3 olume c) Irreversible expansion steps P 3 3 olume w = p 3 ( 3 ) p ( 3 ) more work than single step expansion d) Reversible expansion olume w = p ext d = pd For an Ideal Gas w = pd = nr d = nr d = nr ln 3

4 Joule Expansion In the expression, du = ( U ) d + ( U ) d, ( U ) is C. Now what is ( U )? Lets construct an experiment to determine ( U ) also known as Π So here we have an adiabatic wall q = 0 and w = 0 i.e. ΔU = 0 So, du = C d + Π d = 0 acuu m i.e. Π = ( U ) = C ( ) U = C η J Now we can determine Π as we can measure the change in as a function of. Joule found this η J to be zero for all the gases. For an I.G. U is only a function of U() = U(0) + 3 R (for a monatomic IG) and hence Joule s results are correct for an IG. However, not so for a real gas (we will see later). Enthalpy, H (a state function) ΔU = q + w = q pδ Δp = q p pδ (constant pressure) C p d + μ dp Δ(U + p) = ΔH = q p Since, H = H(p, ) we can write dh = ( H ) d + ( H ) dp = p p he heat capacity at constant pressure, C p = ( H ) p he isothermal Joule-homson coefficient, μ = ( H ) is p determined by the Joule-homson experiment. Conditions of the J- experiment a) Adiabatic b) w = p 1 1 p And ΔU = q + w = 0 + p 1 p = Δ(p) herefore, Δ(U + p) = 0 = ΔH i.e. Constant enthalpy experiment 4

5 dh = C p d + ( H p ) dp = 0 ( H p ) = C p ( p ) = C p μ J H Change in temperature with change in pressure can be measured. For IG: μ J = ( ) = 0 p H For van der Waals gas: ( H p ) b a R Which gives, μ J = b 0 i.e. the inversion temperature, inv = a Rb a = R inv a a If < b i.e. > R Rb ( Δ ) < 0 i.e. gas heats on expanding. Δp H A positive μ implies a cooling on expansion. Principle: Gas expands molecules move apart but are attracted to each other hence lose some KE slow down cool down. his is true when attractive interactions are dominant. Adiabatic expansions a) Ideal gas, reversible Given dq = 0 (adiabatic) dw = pd (reversible) du = C d (I. G. one mole) du = pd (First Law) his gives, C d = pd = R d ) On integrating, = ( 1 R C,, ) C p C = R and therefore = ( 1 Cp C C = ( 1 )γ 1 where γ = /C p 5

6 Which gives, γ 1 = constant (for 1 mole gas) Since γ 1 is always positive, adiabatic expansion gas cools. Should be expected because gas does work and no heat is exchanged. So internal energy must decrease implying a decrease in. Since = P/R we can rearrange P γ = constant Isotherm: P = constant Adiabat: P γ = constant Since γ > 1 at the same pressure, adiabatic expansion gives a lower volume adiabat isotherm olume Irreversible Adiabatic Expansion dq = 0 dw = p d du = C d = p d Which gives, 1 (C + R) = (C + p p 1 R) the gas still cools on expansion as 1 > Now, since (w rev ) > (w irr ), which one gets us to a lower, irr or rev adiabatic expansion? What would happen if we carried out irr adiabatic expansion against vacuum? 6

7 State functions and exact differentials State functions: depend on the state and not how the state has been formed. e.g. internal energy, enthalpy. Does not matter how I reach the state. Whether I change first and hold the constant and then change at constant or I change at constant and then change holding the constant, I end up at the same value of U. i.e. U = U Path functions: are not for the state but for the way the state has been achieved. e.g. work, heat 7