2. m neglect weights of pan, piston and oil Fig1-2 at p7 A h A P

3. Ex 1.2 Piston- Cylinder assembly, diameter = 1cm, mass = 1.64 kg, g = 9.82 m/sec 2, Barometric pressure =748 torr, Absolute pressure = ? F = m g = 60.295 N Gauge pressure = F/A =76.77 N/cm 2 Absolute pressure = 76.77 N/cm 2 + 748 torr (0.01332 N/cm 2 / torr) =86.74 N/cm 2 = 867.4 k Pa Ex 1.3 Mercury manometer reading = 60.5 cm, g = 9.784 m/sec 2 at 27 o C, pressure = ?  = 13.53 g/cm 3 P =  g h = (13.53 g/cm 3 ) (9.784 m/sec 2 ) (60.5 cm) = 8,009 (g m) / (sec 2 cm 2 ) =8.009 N/ cm 2 = 80.09 kPa = 0.8009 bar Unit Conversion : Appendix A Refer to Chemical Engineering Calculation Apply same numerator/denominator to convert the unit g  kg : 1 = 1 kg / 1,000 g torr  bar : 1 = 1 bar / 750.061 torr Complicated systems for students in tests !!!

4. Integration : lose = , gain =  system gains work, W > o  - sign V1tV1t P2P2 P1P1 P V V2tV2t 1 2

6. 1 2 3 4 Ch 1 HW : 3, 5, 6, 7, 12, 13, 15, 18, 19, 24