Question

A heat engine using a monatomic gas follows the cycle shown in the pV diagram. P11 The gas starts out at point 1 with a volume of V1 = 318 cm", a pressure of p1 = 147 kPa, and a temperature of 317 K. The gas is held at a constant volume while it is heated until its temperature reaches 455 K (point 2). The gas is then allowed to expand adiabatically until its pressure is again 147 kPa (point 3). The gas is maintained at this pressure while it is cooled back to its original temperature of 317 K (point 1 again). p (kPa) Adiabat P - V (cm) For the first stage of this process, calculate in joules the heat Q12 transferred to the gas and the work W12 done by the gas. Q 12 = 30.5 J W12 = 0 For the second stage, calculate the heat Q23 transferred to the gas and the work W23 done by the gas. Q23 = 0 J W23 = 36 For the third stage, calculate the heat Q31 transferred to the gas and the work W31 done by the gas. | Q = 116.9 W31 = -46.7

Answer 1

Solution monoa tomwc Яob T,31K 31g c3 V4 Cv 12.5 mal K for process 1 -2 - far monoatomc iocharic Y .66 process r adiabatc Paoces momo Werk done = Pav atomc g So, 8y i*tlau fhermodyna mi es U & AU ncv AT L12.5)(T. -1) 12.5 (455-317) 1725 J 175 Aheat tousfered to adiabatc pro leas o, P vCoushant no heat tog -fe taug plales we have to find v, so,

now, Р. V - R T 3 Р. 318 х 10 а 1.89 х 10 1890 к Ра 3 в hu8u. 8 Cmi now, 1 (В% Вyg) -y (143 х иевея 1890х 318 -66 423.330 Texpansion so, w,=tve. ) fai prоCess S-1 t so charic proceas , Сompveasim do - ve 143 (V,- V) х10 'xi6 -е и (318 - 484.8) х103я108 612.51 J 31 Know now wе 3PVR T У4ох ииия В- до +w лU -п Судт 13S T-7) dоE 12.5 (312-35.2) до - 8.31h х10 2972.5 J now В 2972.5- 612.5 13 Ве 2559491JVs