Question

A heat engine using a monatomic gas follows the cycle shown in the pVpV diagram.

2 Р2 Adiabat PA 1 V. V (cm2) p (kPa)

The gas starts out at point 1 with a volume of V1=233 cm3,V1=233 cm3, a pressure of p1=147 kPa,p1=147 kPa, and a temperature of 317 K.317 K. The gas is held at a constant volume while it is heated until its temperature reaches 395 K395 K (point 2). The gas is then allowed to expand adiabatically until its pressure is again 147 kPa147 kPa (point 3). The gas is maintained at this pressure while it is cooled back to its original temperature of 317 K317 K (point 1 again).

For the first stage of this process, calculate in joules the heat Q12Q12 transferred to the gas and the work W12W12 done by the gas.

Q12= J

W12= J

For the second stage, calculate the heat Q23Q23 transferred to the gas and the work W23W23 done by the gas.

Q23= J

W23= J

For the third stage, calculate the heat Q31Q31 transferred to the gas and the work W31W31 done by the gas.

Q31= J

W31= J

Answer 1

Pokpa) monoatomic gu R 2. 2 (72=395k V-233em T2=395k [8=5/3 L = P2 = 147 kPa) PEL 3 Pointi: V = 23cm² = 233x10 m ² P1 = 147 kPa = 147X103 Pa T1 = 317 K By ideal gas equation, PV-ART 14781038233x108=nX 8.314x317 = n = 34251x103 36355538 0 first stage (142) Cisochonic process) . WH2= POV=P(VF Vis [1412 = o( Vf=vi) Q12 = new T2-Ty) = n. 3R (Tz-Tz) <?x-34251010248-314(395-317) -2 2635538 [812 = 12.64 J Ares 3 is adiabatic process, so 1823=0) Ang Was = nRTs in (N 2 ) = nRTs in (P2) - By Pvz nRT2 > P2= 34251x10 3 x 8.3148395 2635-538 233x106 - 183.2 x103 pa Pz= 183.2 kPa 2

in process 2-3 y (183-2) 395 = (183.3) 73 = 36117k] - from @ H23 = NX 8:31Mx 3617 ln (183-2) [12/23 = 8.60 J Am 831= nCp (T-T3) = 0.013XS38:314x (317-361.7) 1831---12:07 JA 431 = P2(V-V3) =147x10°(233-13) ENR(Ti-T3) 20.013781314 ( 317-361.79 1431²-4.83J . Ang