System 1 P 1 = 6.0 atmV 1 = 0.4 L P ext = 1.5 atm, How much work will be done as the gas expands against the piston? P 2 = V 2 = P 1 V 1 = P 2 V 2 (6.0 atm) 1.6 L w = -P ext  V -(1.5 atm)-1.8 L atm (-1.8 L atm) PV = nRT 1.5 atm (101.3 J/L atm) =-182 J w = (0.4 L)= (1.5 atm)(V2)(V2) ( )L= T = 298 K

System 2 How much work is done when the stopcock is opened? P 1 = 6 atm V 1 = 0.4 L T 1 = 298 K P 2 = V 2 =1.2 = T 2 P 1 V 1 = P 2 V atm w = -P ext  V = -(0 atm) w = L vacuum 0.4 L ideal gas 6.0 atm (1.6 L L) L= 1.6 L

System 1System 2 P 1 = P 2 = V 1 = V 2 = T 1 = T 2 = Same initial and final conditions   E will be the same Ideal gases :  E = q + w = 0 w = -182 Jq = +182 Jw = 0q = 0 all State functions are the same 6.0 atm 0.4 L 298 K atm 1.6 L if  T = 0,  E = P 1 = P 2 = V 1 = V 2 = T 1 = T 2 =

Thermochemistry State 1 = State 2 =reactantsproducts  E = E products - E reactants Endothermic reaction q a) < b) > c) = reaction lowered T of system w 0 0 Ba(OH) 2 8H 2 O (s) 2NH 3 (g) + 2NH 4 SCN (s)  + Ba(SCN) 2 (l)+ 10H 2 O(l) a) < b) > c) =

Endothermic Reaction T prod < T react K.E. prod < K.E. react P.E. prod > P.E. react products less stable than reactants

Exothermic Reaction T prod > T react K.E. prod > K.E. react P.E. prod < P.E. react products more stable than reactants q 0 w 0 < < C 12 H 22 O 11 (s)+ KClO 3 (s)  mix of gases

Calorimetry How is heat measured? It isn’t Temperature measured  T (K) C = heat capacity(J/K) = q (J) C = heat to raise T 1 o x specific heat capacity gJ / K molar heat capacity mol J / K

heat capacities substance specific m.w.molar heat capacity (g/mol) heat capacity (J / K g) (J / K mol) Al (s) Fe (s) H 2 O(s) 36.5 H 2 O(l) CCl 4 (l)

Calorimetry 1. Measure  T (T final - T initial ) 2. Convert to q  T (K) x mass (g) =x C (J/K g) q (J) q is a path function  E = q + w q = EE EE - w + P ext  V

q =  E + P ext  V H q =  E + P  V At constant P,  H =  E + P  V q EnthalpyH  H =  E+ P  V+V  P  P = 0  H =  E +  PV  E + PV p = H =  H q v = E =  E

gases E= K.E. trans + K.E. rot + K.E. vib + P.E. IMF + P.E. bond q =C T T =  H TT TT TT C P = qPqP =  E TT TT C V = qVqV HH =  E +  PV TT TT C P =C V + RTRT TT = C V + R C V =3/2 R C P = 5/2 R Ideal

1000 J of heat 2.00 mol Ar T 1 = 298 K find T 2 at constant volume at constant pressure C V = 3/2 R C P = 5/2 R 1000 J 3/2 x x x(T 2 –T 1 ) 1000 Jx 5/2 x x x(T 2 –T 1 ) = 2 x = 2 T 2 = 338 K 322 K  E = n C V  T = 2 x 3/2 R x = 1000 J  E = n C V  T = 2 x 3/2 R x = 600 J find  E heat and ideal gases

P V AC B D I IIIII IV I q P =P = 5/2R (T C -T A )= 10.1 kJ w = kJ IIq 3/2R (T B -T C ) = kJ w = -P ext  V = 0 kJ V =V = IIIq V =V = 3/2R (T D -T A ) = kJ w = -P ext  V = 0 kJ IV q 5/2R(T B -T D ) = 5.07 kJ w = -P ext  V = kJ P =P =  E = 1.52 kJ  E = 1.52 kJ -P ext  V =  E = 3/2R (T C -T A ) = 6.07 kJ  H = 5/2R (T B -T C ) = kJ  H = 5/2R (T D -T A ) = kJ =  H =  E  E = 3/2R (T B -T D ) = 3.04 kJ =  H  H = 2.51 kJ  H = 2.53 kJ