(i) State the first law of thermodynamics. As ideal gas of volume 6.0 L was made to expand at constant temperature and pressure of 2 atm by supplying heat. If the final volume of the gas was 12.0L, calculate the work done and the heat supplied in joule in the process. [1L*atm=101.3]. (ii) At 0^(@)C H_(2)O(s)hArrH_(2)O(g),DeltaH=51885J*mol^(-1) H_(2)O(l)hArrH_(2)O(g),DeltaH=45860J*mol^(-1) Calculate the change in entropy for the process, H_(2)O(s)toH_(2)O(l) at 0^(@)C.

(i) State the first law of thermodynamics. As ideal gas of volume 6.0

1.	(i) State the first law of thermodynamics. As ideal gas of volume 6.0 L was made to expand at constant temperature and pressure of 2 atm by supplying heat. If the final volume of the gas was 12.0L, calculate the work done and the heat supplied in joule in the process. [1Latm=101.3]. (ii) At 0^(@)C H_(2)O(s)hArrH_(2)O(g),DeltaH=51885Jmol^(-1) H_(2)O(l)hArrH_(2)O(g),DeltaH=45860J*mol^(-1) Calculate the change in entropy for the process, H_(2)O(s)toH_(2)O(l) at 0^(@)C.
Answer» SOLUTION :(i) We know, `w=-P_(ex)(V_(2)-V_(1))` `therefore w=-2(12-6)Latm=-12Latm=-1215.6J` As the PROCESS is ISOTHERMAL and the system is an ideal GAS, `DeltaU=0` for this process according to the 1st law of thermodynamics, `DeltaU=q+w` `therefore 0=q-1215.6 or, q=+1215.6J` (ii)