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 of electrical energy and delivers 150 J of pressure-volume work against the surroundings while releasing 300 J of heat energy. What is the change in the internal energy of the system?
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`
Topper's Solved these Questions
THERMODYNAMICS
NARENDRA AWASTHI|Exercise Level 1 (Q.1 To Q.30)|6 Videos
THERMODYNAMICS
NARENDRA AWASTHI|Exercise Level 1 (Q.31 To Q.60)|2 Videos