In a thermodynamic process the pressure of a fixed mass of a gas is changed in such a manner that the gas molecules give out 30 J of heat and 10 J of work done on the gas. If the initial internal energy of the gas was 40 J. Then the final internal energy will be ?

first law of thermodynamics is dQ=dU+dWdQ is heat energy =-30J  (-ve since heat given out by system)dW is work done = -10J (-ve since work is done on system)dU is change in internal energy=Uf-Uii=Ufinal-Uinitial here Ui=40J  substitute the values in first law of thermodynamics  dQ=[Uf-Ui]+dW ⇒-30=[Uf−40]−10 ⇒-30+50=Uf ⇒Uf=20 J