The first law of thermodynamics for a closed system is dU = dq + dw, where dw = `dw_(pv)+dw_("non-pv")`. The most common type of `w_("non-pv")` is electrical work. As per IUPAC convention work done on the system is positive. A system generates 50 J electrical energy, has 150 J of pressure-volume work done on it by the surroundings while releasing 300 J of heat energy. What is the change in the internal energy of the sytem?
A
`-500`
B
`-100`
C
`-300`
D
`-200`
Text Solution
Verified by Experts
The correct Answer is:
D
`DeltaU=q+w` `DeltaU=-300+(-50+150)=-200`
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