Question

Please help me about Physics, Thanks. A sample of 1.00 mole of a diatomic ideal gas is intially at temperature 265K...........

Thermodynamic Processes involving Ideal Gases-in-class worksheet-(5 points) PHYS 181 Question B (B.) A sample of 1.00 mole of a diatomic ideal gas is initially at temperature 265 K and volume 0.200 m. The gas first undergoes an isobaric expansion, such that its temperature increases by 120.0 K. It then undergoes an adiabatic expansion so that its final volume is 0.360 m. i. Sketch a PV diagram for the two-step process, including labeled initial, final, and intermediate (after the isobaric process, before the adiabatic process) states, and a two-part curve/path with an arrow indicating direction. Label the initial state "1", the final state "f", and the intermediate state "b". Write down the known values for P, T, and V at each point, e.g. Ti = 265 K, and 7) = 385 K. 3. (0.7 points) What is the initial pressure of the gas, P, in kPa? 4. (1.1 points) What is the total heat transfer, Q, to the gas, in J? 5. (1.5 points) What is the total work done on the gas, W, in J? (Enter your answer for problem (5.) for credit. First, use the following questions as intermediate steps; answers can be submitted to assess understanding of the problem along the way.) i. What is the work done on the gas in just the isobaric process, Wisobaric in J? ii. What is the intermediate volume of the gas, V. (after the isobaric process) in m3? ii. What is the final temperature, Ty of the gas in K? Use the known values of VT, and V (from part 5ii.), along with the relationship between temperature and volume for an adiabatic process iv. What is the work done on the gas in just the adiabatic process, Wadiabatic, in J? Check: Consider the First Law of Thermodynamics for this entire process. Does AE int = nC,(T; - Ti) = (total + Wtotal?

Answer 1

Here we apply definition of thermodynamics processes and first law of thermodynamics and ideal gas law as well as definition of internal energy and work.

F mol Pi = Pb (as isobaric) Vj = 0.200 m3 n = 1 mole diatomic gas af (p = 7 R, Cu = SR. - Ti = 265 K Temperature at b' T = 385k . 3 Using ideal gas law L_Pi Vi = DR Ti sy Pi (0:200 m2 )= (1 mol ) (83143) (265K) z Pi = 11,016:05 Pascal so cinitial Pressure is 11,016.05 Pascal. 4) for adiabatic process, we know na heat exchange takes place. - so Qbf so for isobaric Process Heat exchanged, Q = Alib Here Sti. = Enthalpy change from jutob

- no va T. a Hip = nCp (To-T;) to Qi = (1) 7 R (ToT) A = (0) 368.3145) (385-265)k = 3,491-88 Joule so total heat transfer to gas. Q = Qio + Q bf = 3491.885 ③ As for itb Pressure is constant su - The 3 Vo = 13850.2m3) = 0.281m3 (265) Work done, Wit = Pi (V-V. =(11,01605 )(0.291-0.200) Joule = 997.685 final volume Vf = 0.360m². Vi = 0.291 m3 : We know b f adiabatic process - PUX = constant ON DRTUA constanta

- Tural = Constant = TE VH = To Vott -= P = 2/3 = 14 > Tg = 385K) (0.291.36*** If = 353.6K 10.360, so change in internal energy for b f SULE = nCv (Tf - !=(1) 5 (8:314)(353-6-385) J = -652.9 II Qbf=0, Was = ? from first law of thermodynamics Qbf = AU of tWof 7 Wof = 0 - (-652.91) - Wbf = 652.915 Total Work Witt = Wiot Wof = 097,68 +652.91)J =1650.595