Question

4

A piston is filled with 75.0 mol of an ideal gas at a pressure of 3.40x10Pa at a temperature of 1250 K. While keeping the pressure constant, the volume of the piston is reduced to 0.530 m. The gas is then allowed to expand slowly while keeping the temperature constant until the volume reaches 4.16 m2 The gas is then compressed rapidly until the gas reaches its original state. Complete the chart below. Do not include units in your answer. State Vim) P(Pa) T(K) Work(J) Heat(J) W30 Qab- A 3.40x105 1250 ΔUO) AU- Wc- Qbc- AU B 0.530 W Qca" AU. C

Answer 1

B - ) - 1 = 75x80314 x 1250 3 Solution)A) number of mol = 75=n. R= 8.314 Paimk mol Initial Temperature 1250K. Initial volume = - RX RX 1250 3.4x10S 3.4 x 105 Initial volume = 2.29 m Wab = - 3:4 x 10S (0:53-2:29) = 5.984x10 J final Temperature = 314x108 8.53 x 8.5=289 K. 75x8.314 75x3x8.314 1250) Şassuming the gas is mono- atomic svab = -898847.325 J s AUab (289 2

3 БС bc. - 75x80314x 289 x In/4.10 ) Gmb = AU - W = -300447•325 J golution B) Initial volume = 0.53 m Temperature 289 R. Press wie = 3.4 x 10 Pa. 0.53 Wr bc = - 371295 •14 : AU= 0 SIsothermal PJTO CO 2 Inc = 371295: 14 J: solution c) Initial volume = Initial pressure 75x8.314 x 289 – 43318.73 Pa 4016 Initial Temperature =289 K. & Iso thermal Proces Wea = 754 - 43318.73 (2029 - 4•10) J Wca =-81006. 04 J AU = 0 {Iso thermal process}. I ca=81006.04J. 4.16m3 - }

State V (m3) P (Pa) TCK) 2.29 3.4x105 1250 Work (J) Heat (J) AULJ) 5.984x165-300447. -89 8847. A 325 325 B 0.53 3.4x10t 289 -371295014 371295.14 0 o C4016 43318.731 289 -81006-0481006.04