First Law: ( \Delta U = Q - W = 50 - 40 = 10 , \textkJ )
: Work done at const. pressure: ( W = P \Delta V = 2 \times 10^5 , \textPa \times (0.3 - 0.1) , \textm^3 ) ( W = 2 \times 10^5 \times 0.2 = 40,000 , \textJ = 40 , \textkJ ) chemical engineering thermodynamics yvc rao pdf 27
: A closed system containing air undergoes a constant pressure process at 2 bar. The volume increases from 0.1 m³ to 0.3 m³. During the process, 50 kJ of heat is added. Calculate the change in internal energy. First Law: ( \Delta U = Q -
Answer: Internal energy increases by 10 kJ. During the process, 50 kJ of heat is added
| Method | Access to Page 27 | Cost | |--------|------------------|------| | Purchase physical copy (Universities Press) | Yes | ~₹500 | | University library | Yes | Free for students | | Google Books (limited preview) | Maybe | Free | | Kindle eBook (Amazon) | Yes | ~₹400 | | Course reserves (institutional login) | Yes | Free | | Inter-library loan | Yes | Often free |
If you’re an instructor, consider putting the relevant pages (e.g., page 27) on your university’s LMS. If you’re a student, buy the book or borrow it — your future engineering ethics will thank you.
