Review: Problem Solving in Isothermal Process 02

(Last Updated On: June 7, 2023)

Problem Statement:

During a reversible process there are abstracted 321 kJ/s from 1.234 kg/s of a certain gas while the temperature remains constant at 29.67°C. For this gas, Cp = 2.232 and Cv = 1.713 kJ/kg.°K. The initial pressure is 687 kPa. For both nonflow and steady flow (ΔP = 0, ΔK = 0) process. Determine (a) v1, v2 and P2 (b) the work and (c) ΔS and ΔH.

Course: Thermodynamics

TOPIC: Processes of Ideal Gas SUBTOPIC: Isothermal Process DEFINITION: An Isothermal Process is an internally reversible constant temperature process of a substance.

Problem Answer:

a.) v1 = 0.2823 m^3/s, v2 = 0.0539 m^3/s, P2 = 3598.1466 kPa b.) Wn = Ws = Q = -321 kJ/s c.) ∆H = 0

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