11.1 1st Law of Thermodynamics A law is a statement which summarizes our experiences. Among the most fundamental laws (there are no known exceptions to this law) is the 1st Law of Thermodynamics which states that the total mass and energy in the universe (an isolated system) is constant.  E universe =  E system  E surroundings  = 0 Why is mass included in this statement? Energy is a state function, i.e. dE is an exact differential, and therefore energy has definite values in the initial and final states of the system:  E = 1  2 dE = E 2 - E 1 For a cyclic process: (state 1) (state 2)  E =  c ycle dE = 1  2 dE + 2  1 dE = (E 2 - E 1 ) + (E 1 - E 2 ) = 0 In general cyclic integrals over state functions are by definition zero and in fact this property is used to define state functions:  d (state function)  0 The circle on the integral sign indicates that the integration is taken over a cyclic path.

11.2 Since changes in energy usually appear in the form of heat or work in a closed system, two other expressions of the 1st Law of Thermodynamics are for differential changes: dE = dq + dw and for finite changes :  E = q + w Why don’t dq and dw integrate to  q and  w for a finite change? For a process carried out at constant volume, dV=0, and in a closed system involving only pressure volume work:  E = q + w = q -  P ext dV = q V - 0 = q V The heat transferred in a process carried out at constant volume in a closed system involving only pressure volume work is equal to the change in internal energy. For finite changes:  E = q V and for differential changes: dE = dq V and since C V = (  q/  T) V = dq V /dT = (  E/  T) V dE = C V dT in a constant volume process in a closed system involving only pressure volume work. = 0

11.3 Calculate the heat required to heat 2.00 moles of a gas at constant volume from 300 K to 600 K? The constant volume heat capacity of this gas is independent of temperature and is 12.5 Joules /mole K. 300 K600 K q V =  E =  dE = 300 K  600 K C V dT = (2.00 mol) (12.5 J /mole K) (600 K K) (1 kJ / 10 3 J) = 7.50 kJ

11.4 Most processes are carried out at constant pressure (usually the ambient barometric pressure) and so the heat transferred in a constant pressure process is of interest. Starting with the 1st Law:  E = E 2 - E 1 = q + w = q -  P ext dV = q P - P ext [V 2 - V 1 ] When the system pressure is constant, the external pressure of the system must equal the system pressure. Furthermore the system pressure is also the pressure in the initial and final state, since the system pressure is constant:  E = E 2 - E 1 = q P - P system [V 2 - V 1 ] = q P - (P 2 V 2 - P 1 V 1 ) Solving this equation for q P gives: (E 2 + P 2 V 2 ) - (E 1 + P 1 V 1 ) = q P which suggests defining a new state function the enthalpy, H, as: H  E + P V With this definition we see that the change in enthalpy,  H, is equal to the heat transferred, q P, in a constant pressure process in a closed system involving only pressure volume work. For finite changes:      H = q P and for differential changes: dH = dq P and since C p = (  q/  T) P = dq P /dT = (  H/  T) P dH = C P dT in a constant pressure process in a closed system.

11.5 P ext = 1.00 atm Calculate the heat required to heat 2.00 moles of a gas at constant pressure from 300 K to 600 K? The constant pressure heat capacity of this gas is independent of temperature and is 20.8 Joules /mole K. 300 K600 K P gas = 1.00 atm q P =  H =  dH = 300 K  600 K C P dT = (2.00 mol) (20.8 J /mole K) (600 K K) (1 kJ / 10 3 J) = 12.5 kJ Why was more heat required to heat the gas from 300 K to 600 K when the pressure was held constant as compared to when the volume was held constant? What is the difference q P - q V equal to (I’m not looking for a numerical answer)?

11.6 The heat transferred between the system and the surroundings in a process depends on the path taken between the initial and final states, but the change in internal energy and change in enthapy, which are state functions do not! The sketch below, hopefully will make these distinctions clearer: (state 1) (state 2) constant volume path (closed system with only PV work): q v =  E =  Cv dT  H =  Cp dT (since H is a state function  H will be calculated the same way, independent of the path taken) constant pressure path (closed system with only PV work): q p =  H =  Cp dT  E =  Cv dT (since E is a state function,  E will be calculated the same way, independent of the path taken)

11.7 The definition of enthalpy can be used to derive a useful relationship between C p and C v for ideal gases: H  E + P V = E + n R T Taking the partial derivative with respect to temperature while holding the pressure constant gives: (  H/  T) P = (  E/  T) P + n R Since for an ideal gas E is a function of temperature only and does not depend on pressure or volume: (  E/  T) P = (  E/  T) V and: (  H/  T) P = (  E/  T) V + n R C P = C V + n R i.e., for an ideal gas the constant pressure molar heat capacity exceeds the constant volume molar heat capacity by the gas constant, R. Comment on how and when this relation would apply to real gases.

11.8 Calculate the internal energy change when moles of O 2 (g) are heated at constant pressure from K to K. The temperature dependent constant pressure molar heat capacity of O 2 (g) is: C P (cal / mole K) = x T x T -2