ThermodynamicsHard
Question
A system absorbs 100 kJ heat in the process shown in the figure. What is ΔU for the system?
Options
A.−50 kJ
B.+50 kJ
C.+150 kJ
D.−150 kJ
Solution
$W = - 1 \times \frac{1 + 2}{2} = - 1.5\text{ bar.}\text{m}^{3} = \frac{- 1.5 \times 1000 \times 100}{1000}\text{ kJ = } - 150\text{ kJ}$
and $q = + 100\text{ kJ}$
$\therefore\Delta U = q + w = - 50\text{ kJ}$
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