In This post you will get answer of Q.
A monoatomic gas supplied the heat $Q$ very slowly keeping the pressure constant. The work done by the gas will be:
Solution:
For monoatomic gas at constant pressure
$ frac{Delta U}{Q}=frac{1}{3} $
or $ Delta U=frac{Q}{3} $
Now applying first law of thermodynamics
$ W=Delta Q-Delta U $
$ =Q-,frac{Q}{3},=frac{2Q}{3} $
Solution:
For monoatomic gas at constant pressure
$ frac{Delta U}{Q}=frac{1}{3} $
or $ Delta U=frac{Q}{3} $
Now applying first law of thermodynamics
$ W=Delta Q-Delta U $
$ =Q-,frac{Q}{3},=frac{2Q}{3} $