2010 Semifinal Exam Part A 6 乃=s(得) PoVo =nRTo Vo=nRTo Po %=1atm=1.01.105Pa 2 mols8.31 molK298K %= 1.01·105Pa %=0.0490m3 4-7%=}00490m3=0123m 1 2=15.0L=15.0L. 103 mL 1 cm3 1 m3 IL 1 mL 106cm3=0.0150m3 T2=298K 0.01231 0.0150 T2=275K d.Lowest Pressure From the P-V diagram,the lowest pressure occurs at the end of the second isothermal process. P3V3=nRT3 A-0 T3=T2=275K 3=%=0.0490m3 P3= 2.00 mol.1 m75K 0.0490m3 3=9.33.104Pa Copyright C2010 American Association of Physics Teachers2010 Semifinal Exam Part A 6 T2 = 298 K  V1 V2 2 5 P0V0 = nRT0 V0 = nRT0 P0 P0 = 1 atm = 1.01 · 105 Pa V0 = 2 mols · 8.31 J mol·K · 298 K 1.01 · 105 Pa V0 = 0.0490 m3 V1 = 1 4 V0 = 1 4 · 0.0490 m3 = 0.0123 m3 V2 = 15.0 L = 15.0 L · 103 mL 1 L · 1 cm3 1 mL · 1 m3 106 cm3 = 0.0150 m3 T2 = 298 K  0.0123 0.01502 5 T2 = 275 K d. Lowest Pressure From the P-V diagram, the lowest pressure occurs at the end of the second isothermal process. P3V3 = nRT3 P3 = nRT3 V3 T3 = T2 = 275 K V3 = V0 = 0.0490 m3 P3 = 2.00 mols · 8.31 J mol·K · 275 K 0.0490 m3 P3 = 9.33 · 104 Pa Copyright c 2010 American Association of Physics Teachers