Question

Goal Solve for the performance coefficient of a refrigerator using a five- step process the includes: 1. Making a state table. 2. Making a process table. 3. Calculating the totals for Work, Heat, and Internal-Energy-Change. 4. Identifying the heat input (cold reservoir) and output (hot reservoir). 5. Calculating the performance coefficient of the refrigerator. isothermal Problem Shown in the figure to the right is a cyclic process undergone by a refrigerator. Your refrigerator shall use 8.0 moles of helium gas (monatomic). During the process a->b, the volume increases by a factor of 3.0. T = 300 K Pa = 100,000 Pa Refrigerator Cycle Solution (1) Fill in the State Table (all pressures in Pascals, all volumes in cubic meters, all temperatures in K). Pressure Volume Temperature Work Heat V (2) Fill in the Process Table (all entries in Joules). b->c C->a (3) Find the Totals: Work = Heat = dU = (4) Find the heat input (from "cold reservoir") and the heat output (to "hot reservoir"): Q-hot = Q-cold = (5) Find the performance coefficient of the refrigerator: Coefficient =

Answer 1

hinen ue ata, pa-lx105 Pa T 300K let vaurmer v Va Va hRTa - '8x'8.314x 300 D-199m n o.2m3 -P v No APaat b 3XVa 3xu:2=Q.6 m 105x06 AoPPa = 105 Pa Tb= 8x6-314 902.09K TTC9 02.09 K e-314 g0 209 hRT V. O 2 3x 1o5 Pa P-3.993 Vaume Tmpersturd Phessure 0-2 m3 (05 Pa S00K O-6m3 J02.09r 3 Pa (05 (a6-o-2) -uxo'T WonP(Vbva ) i0 (es ab, aUan- T-6-7-1 AUab x2-u (902-300) - 52K 59.7249z 99.72K2 Auay eWop Heat d

PioCe bac W nem BxB3102 Trbc_65.9%J (OT20 PrOasn C-a, Ta-te)= Bx12-4 (300-902) 7-59.72 K- AUCat Wca Heat 59.72 (a-1) 99 72 -65.9 65.9 -59.72 -59.72 = -25.9 kg 65.9 uo- Tide Woik= 99-72-65g -59.72 - -25.9 ky Total Het- S.62 93.72 K: 99.72 KT 99.12 Paformnc CoP (25.62-99-72 Cops 3.05