# 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.

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
View Solution:

Subscribe To Unlock The Content! and Remove Ads.

More Questions in: Thermodynamics Problems

#### Online Questions and Answers in Thermodynamics

P inoyBIX educates thousands of reviewers and students a day in preparation for their board examinations. Also provides professionals with materials for their lectures and practice exams. Help me go forward with the same spirit.

“Will you subscribe today via YOUTUBE?”

Subscribe

• Full Content Access Exclusive to Premium members

## PINOYBIX FREEBIES FOR PREMIUM MEMBERSHIP:

• CIVIL ENGINEERING REVIEWER
• CIVIL SERVICE EXAM REVIEWER
• CRIMINOLOGY REVIEWER
• ELECTRONICS ENGINEERING REVIEWER (ECE/ECT)
• ELECTRICAL ENGINEERING & RME REVIEWER
• FIRE OFFICER EXAMINATION REVIEWER
• LET REVIEWER
• MASTER PLUMBER REVIEWER
• MECHANICAL ENGINEERING REVIEWER
• NAPOLCOM REVIEWER

## FOR A LIMITED TIME

If you subscribe for PREMIUM today!