Question

Isothermal b Adiabatic

7.5) A 1.15 -mol quantity of monatomic ideal gas undergoes the following cyclic process. The gas starts at point a at STP. It expands isothermally to point b, where the volume is 2.2 times its original volume. Next, heat is removed while keeping the volume constant and reducing the pressure. Finally, the gas undergoes adiabatic compression, returning to point a.

a. Calculate the pressures at b and c. (answers in Pa)

**Find the volumes at a and b first.
**Use the relations between pressure and volume for isothermal and adiabatic processes, as appropriate.Use the relation between the volume and temperature for an adiabatic process.

b. Determine the temperature at c. (answer in kelvin)

c. Find the work W done by the gas, heat Q added to the gas, and the change in entropy ΔS of the gas in each step (a-b, b-c, c-a) --> (expecting 9 answers)

**Since T is constant in an isothermal process, ΔS = Q/T

**Use the relation dS = dQ/T, and note that dQ = dU for this process. Hence dS = (3/2) n R dT/T.

d. Calculate the efficiency e of this cycle.

Answer 1

Solution Isothemal A diabahe Cu = 3 R Mono atomic : a. For process ab Pava = Po Vo Va Pax Poz Pa 2-2Va Pb= Pa O.454 atm 0.459 x 105 Paseals Pa Vas nR Ta At STP, Ta-273K nRTa Pa Vaz I15 x 8.314x273 1.013 x105 Va - 0.026mo

At b Voz2.2Va Voz 2-2x 0.026 Vc 0.0572 m3 for Pc Ve = Pa Va adiabatic process Va Pc= Pa Pcz Va 2-2 Va Pc=! 0: 268 atm 3.72 0.2715 x10 Pascals 1 atm= 1.013x105 Pascals b. Temperature at c PcVc 0.271sX16x0.052 1.15 x8.314 %3D Tc = 162.43 K

Vo Was c) Va Way= l15X8.314x273 lu nR Ta Ju ( c) Wab ) 1.15x8.314 x fu(2-2) x 273 Wab= X273 dab Wab= "7.54.I Wab- 2058.42 J Wbe = ( const volume þrocess) Adbc= AUbc- nCv ( Tc -Tb) = 1.15x 2R(162-43-273) >. \५.३५16F* -110·S? nR [ Ta -Tc] Wea = 1-r Weaa LISxB:31iu [ 273 -162-43] 1.15x8.314 -1585.76J Wea = l057.2 -2 13

Entropy:- for process T= cmst. 7. 54 J/K 2058.42 273 V= cmstaut, dd = du for process be Te dT S= ds- Ta TC S= ? x1:15*8.318 글지 162.43 273 S= -14.34165x(.52) - 1.S५ J | for C-a , dd=0 adiabatic process !. ds = O

d. Etticienay 1585.76 2058.42- %3D 2 058.42 02 Etticiency= 2058.42 0.2296 N= 22.96%- or 23/- 318