Question

Question: The PV plot (Figure 1) depicts 58.8 mols of a gas with f = 3 going from state A with PA = 1.22 x 10 Pa and VA = 0.785 m going to state B with VB = 3.72 m. Figure 1: A PV plot depicting the described gas going right from state A to state B via a horizontal pathway. The x-axis represents volume and the y-axis represents pressure. Part 1) What is the temperature of the gas at state A? TA = Part 2) Calculate the work done on the gas as it goes from A → B. WA+B= Part 3) Calculate the change in internal energy as the gas goes from A → B. AEint = Check

Answer are:

1. 196

2. -3.58*10^5

3. 5.37*10^5

Please provide the working out. thanks

Answer 1

11 using ideal gas law, PAVA= nRTA - TAPANA Gasconstant R=8.314 Jomal kt . MR To = 1.22x10x 0.785 = 1966 58.8 x8.314 2) Work done w a-POV for constant pressure W=-PA (verv =-1.22x18 (3.72-0.785) W=-3.58x105 7 3) Firal temperature To =P VB = 1.22x10x3.72 - 928-36K R 58.8x8.314 Heat change (0 = CouT= (F+1) NRAT = 3 x 58.8x8. 20 = 8.95x15 J Chonge in internal energy au=CO+OW = 8.95 -3.58x10 XU=5.37x10s ]