Question
If the heat of 110 J is added to a gaseous system, whose internal energy is 40 J, then the amount of external work done is
180J
70J
110J
30J
Solution
B.
70J
Heat added to the system ΔQ =110J
Internal energy Δu =40J
External work done is given by
ΔW= ΔQ-Δu=110-40=70J
