1. A refrigerator operates in a reversed Brayton cycle using 0.0070 moles of amonatomic ideal gas. The gas starts at a pressure of 150 kPa and a temperature of200 K. The gas is then expanded at constant pressure to a temperature of 250 K.The gas is compressed adiabatically to a pressure of 750 kPa. The gas is thencompressed at constant pressure to a temperature of 381 K. The gas is thenexpanded adiabatically to its original state.a. Sketch this cycle on a PV diagram.b. Calculate the heat removed from the cold reservoir during each cycle.с.Calculate the heat expelled to the hot reservoir during each cycle.d. Calculate the COPR for this refrigerator.e. Calculate the power consumption (in Watts) for this refrigerator.Pa.P Vq =nRTq → (1.48atm) Vq = (0.007 mol)(0.05ao6atm) (200k) → V½= 0.078LL. atmmol.k85 =nRT, → (1.45 atm) =(0007 0$)(0.05206 atn)(asok) → Va: 0.0 47L750kPaP,v = P, v = (146atm)(o.0 97L) = (7.40 ata) Vs → Vz= 0.03L1.41.4→ V3= 0.03L1SO fat+0.078L0.097L

Answered: 1. A refrigerator operates in a…

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Chapter1: Introduction

Section: Chapter Questions

Problem 1.1P

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1. A refrigerator operates in a reversed Brayton cycle using 0.0070 moles of a
monatomic ideal gas. The gas starts at a pressure of 150 kPa and a temperature of
200 K. The gas is then expanded at constant pressure to a temperature of 250 K.
The gas is compressed adiabatically to a pressure of 750 kPa. The gas is then
compressed at constant pressure to a temperature of 381 K. The gas is then
expanded adiabatically to its original state.
a. Sketch this cycle on a PV diagram.
b. Calculate the heat removed from the cold reservoir during each cycle.
с.
Calculate the heat expelled to the hot reservoir during each cycle.
d. Calculate the COPR for this refrigerator.
e. Calculate the power consumption (in Watts) for this refrigerator.
P
a.
P Vq =nRTq → (1.48atm) Vq = (0.007 mol)(0.05ao6atm) (200k) → V½= 0.078L
L. atm
mol.k
85 =nRT, → (1.45 atm) =(0007 0$)(0.05206 atn)(asok) → Va: 0.0 47L
750kPa
P,v = P, v = (146atm)(o.0 97L) = (7.40 ata) Vs → Vz= 0.03L
1.4
1.4
→ V3= 0.03L
1SO fat
+
0.078L
0.097L

Expert Solution

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NOTE: Because you have posted a question with multiple subparts, we will solve the first three for you. If you want the other subparts to be solved, please repost the question and specify the subparts to be solved.

Given:

n= 0.007mol

Because the gas is monoatomic, C_p = 5R/2 and C_v = 3R/2

$γ = \frac{C_{p}}{C_{v}} = \frac{5 / 2}{3 / 2} = 1.67$

Formula Used:

PV=nRT,

PV^$γ$= Constant [for adiabatic process]

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