A sample consisting of 2.5 mol of perfect gas molecules with Cp,m = 29.4 J K–1 mol–1 is initially at 303 kPa and 335 K. It undergoes reversible adiabatic expansion until its pressure reaches 250 kPa. Calculate the final volume and temperature and the work done. Answer: w = -900 J

A sample consisting of 2.5 mol of perfect gas molecules with Cp,m = 29.4 J K–1 mol–1 is initially at 303 kPa and 335 K. It undergoes reversible adiabatic expansion until its pressure reaches 250 kPa. Calculate the final volume and temperature and the work done. Answer: w = -900 J

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter12: Thermodynamic Processes And Thermochemistry

Section: Chapter Questions

Problem 21P

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A sample consisting of 2.5 mol of perfect gas molecules with
Cp,m = 29.4 J K^–1 mol^–1 is initially at 303 kPa and 335 K. It undergoes reversible adiabatic expansion until its pressure reaches 250 kPa. Calculate the final volume and temperature and the work done. Answer: w = -900 J

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PHYSICALCHEMISTRY_SHOWSOLUTION:A sample consisting of 2.5 mol of perfect gas molecules with Cp,m = 29.4 J K –1 mol–1 is initially at 303 kPa and 335 K. It undergoes reversible adiabatic expansion until its pressure reaches 250 kPa. Calculate the final volume and temperature and the work done. ANSWER:[w = -9.00x102 J]
Calculate the work done by a gas w hen it expandsthrough (a) 1.0 cm3, (b) 1.0 dm3 against an atmospheric pressure of 100 kPa . What work must be done to compress thegas back to original state in each case?
When a sample of 1.0 mol of a monatomic perfect gas with Cv,m= 12.47 J/mol K and initially at 25°C, doubles it volume adiabatically its temperature decreases to -60°C. State whether the process is reversible or not. Why?
(b) An ideal gas of 2.5 moles undergoes an adiabatic expansion at an initial pressure of 4.25 bar and at an initial temperature of 540 K to a final temperature of 400 K. Calculate the work done which is equal to internal energy change under such conditions. Assume Cp,m = 5/2 R