In This post you will get answer of Q.
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:
Solution:
Heat added to the system $ Delta Q=110 J $
Internal energy $ Delta U=40 J $
External work done is given by
$ Delta W=Delta Q-Delta U=110-40 $
= 70 J
Solution: