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Answer: Internal energy increases by 10 kJ. chemical engineering thermodynamics yvc rao pdf 27
: 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 ) | Method | Access to Page 27 |