Volume Work Calculation of volume Work(i.e. Expansion work, Compression work) Case 1: Expansion or Compression under a constant external pressure, Pex Initial height: Final height Piston movement:△L F F L2 W-F: AL;(Force Distance) F A W=P·A·△L Bottom Surface area=A W=P·△V
Volume Work Calculation of Volume Work (i.e. Expansion work, Compression work) Case 1: Expansion or Compression under a constant external pressure, Pex) 11 Initial height： L1 Final height： L2 Piston movement： L W = F · L; (Force ·Distance） F = Pex·A W = Pex ·A · L Bottom Surface Area = A W = Pex· V
Volume Work Calculation of volume Work(i.e. Expansion work, Compression work) Case 2: Expansion under vacuum, Pex =0) Initial height Final height: F Piston movement:△L F Lw=F: AL,(Force Distance ) W=P…·△V=0 Bottom Surface area=A
Volume Work Calculation of Volume Work (i.e. Expansion work, Compression work) Case 2: Expansion under vacuum, Pex = 0) 12 Initial height： L1 Final height： L2 Piston movement： L W = F · L; (Force ·Distance） W = Pex· V = 0 Bottom Surface Area = A
Practice Q2 Inside a container of 22.4 dm3. there are some amount of ideal gas, with a total pressure of 1.0 atm(1.0 atm=101 kPa). Under a constant temperature T=273 K, the gas expands against a constant external pressure of 0.5 atm How much work will the system do to the surroundings? Solution State 1(1.0 at, 22. 4 dm, n, 273 K)> State 2(0.5 atm, V2, n, 273 K) Moles of gas: n= PV/RT=1.0X 22.4/(0.082X273)=1 mol Final gas volume: V2=nRT/P2=nRT/Pex=10.082273/0.5=44.8 dI Work done by system: W=Pex*(v2-V1=0.5*(448-22. 4) 11.2atm·dm3=1133J (Note: 1 atm =101 kPa, 1J=1Pa*1 m3= 1 kPa*1 L)
Practice Q2: Inside a container of 22.4 dm3 , there are some amount of ideal gas, with a total pressure of 1.0 atm (1.0 atm = 101 kPa). Under a constant temperature T = 273 K, the gas expands against a constant external pressure of 0.5 atm. How much work will the system do to the surroundings? Solution: State 1 (1.0 atm, 22.4 dm3 , n, 273 K) State 2 (0.5 atm, V2 , n, 273 K) Moles of gas: n = PV/RT = 1.0×22.4/(0.082×273) = 1 mol Final gas volume: V2 = nRT/P2 = nRT/Pex = 1*0.082*273/0.5 = 44.8 dm3 Work done by system: W = Pex * (V2 – V1 ) = 0.5* (44.8 – 22.4) = 11.2 atm ∙ dm3 = 1133 J (Note: 1 atm = 101 kPa, 1 J = 1 Pa * 1 m3 = 1 kPa * 1 L) 13
Practice Q3: At 1.0 atm, 100 oC, 1 mol of liquid water evaporates into 1 mol of steam calculate the expansion work done by the system Solution WEP. AV e eX 1 atm X △V=V gas v liqui 9as-0 gas=nRT/P=1×0.082×373.1511.0=306L W=Pex Vaas =1.0 atm X 30.6 L=30.6 atm . L=3100 J
Practice Q3: At 1.0 atm, 100 oC, 1 mol of liquid water evaporates into 1 mol of steam. Calculate the expansion work done by the system. Solution: W = Pex·V Pex = 1 atm V = Vgas – Vliquid Vgas – 0 = Vgas Vgas = nRT / P = 1 × 0.082 × 373.15 / 1.0 = 30.6 L W = Pex·Vgas = 1.0 atm x 30.6 L = 30.6 atm ∙ L = 3100 J 14
Enthalpy Enthalpy(H) H=U+PV and both u and h are state functions We never know the absolute value of u. or h Therefore, the important thing is the change of U or H Enthalpy Change(AH= Hinal-Hinitial) The thermal energy gained or lost under a constant pressure, (applicable for most of chemical reactions) Thermo-chemical equation: eg.CH4(g)+202(g)→CO2(9)+2H2O()+890.3kJ AH=-8903 KJ, and it is for 1 quantity of this reaction. 15
Enthalpy Enthalpy (H): H = U + PV, and both U and H are State Functions! We never know the absolute value of U, or H. Therefore, the important thing is the change of U or H. Enthalpy Change (ΔH = Hfinal – Hinitial) The thermal energy gained or lost under a constant pressure, (applicable for most of chemical reactions) Thermo-chemical equation: e.g. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) + 890.3 kJ ΔH = – 890.3 kJ, and it is for ―1 quantity‖ of this reaction. 15