1 "cm"^3 of water at 1 atm pressure is converted into steam, volume of steam becomes 1671 "cm"^3 hence increase in internal energy will be ____

1 "cm"^3 of water at 1 atm pressure is converted into steam, volume of

1.	1 "cm"^3 of water at 1 atm pressure is converted into steam, volume of steam becomes 1671 "cm"^3 hence increase in internal energy will be ____
Answer» 2087 J 167 J 373 cal 373 J Solution :`DeltaQ=mL_V` =1X 540 =540 cal Now from the first law of THERMODYNAMICS, `DeltaQ=DeltaU+DeltaW` `DeltaQ=DeltaU+PDeltaV` `THEREFORE DeltaU=DeltaQ-PDeltaV` `=540-10^5(1671-1)xx10^(-6)` `=540-1670xx10^(-1)` =540-167 =373 cal