One mole of perfect gas undergoes a reversible compression at constant pressure when the temperature is changed from 310 K to 270 K. Calculate w, q and AU assuming the Cp.m of the gas equal to 29.8 J/mol.K.
Added by Jonathon M.
Step 1
Step 1: Determine the change in temperature (ΔT) Given: Initial temperature (T1) = 310 K Final temperature (T2) = 270 K ΔT = T2 - T1 ΔT = 270 K - 310 K ΔT = -40 K Show more…
Show all steps
Your feedback will help us improve your experience
Kajal Raghubanshi and 77 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Three moles of an ideal gas undergo a reversible isothermal compression at $20.0^{\circ} \mathrm{C}$ . During this compression, 1850 $\mathrm{J}$ of work is done on the gas. What is the change of entropy of the gas?
Kajal R.
The temperature of 1.75 moles of an ideal gas increases from $10.2^{\circ} \mathrm{C}$ to $48.6^{\circ} \mathrm{C}$ as the gas is compressed adiabatically. Calculate $q, w, \Delta U,$ and $\Delta H$ for this process, assuming that $C_{V, m}=3 R / 2$
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD