Correct condition for vaporisation of water at 373 K and 1 atm pressure is `square Delta H>0 ` `square Delta S<0` `square Delta G>0` `square Delta T>0`

For the vaporisation of water : H_(2)O (I) hArr H_(2)O(g) [1 atm. Pressure] Given : Delta S = 120 JK^(-1) and Delta H =+45.0 kJ . Evaluate the temperature at which liquid water and water vapour are in equilibrium at 1 atm. Pressure-

Assertion: The increase in internal energy (DeltaE) for the vaporisation of 1 mole of water at 1 atm and 373K is zero. Reason: For all isothermal processes DeltaE=0 .

The enthalpy change for the reaction of 5 litre of ethylene with 5 litre of H_(2) gas at 1.5 atm pressure is Delta H= -0.5 kJ . The value of Delta U will be: ( 1 atm Lt=100 J)

Assertion:The increase in internal energy (DeltaE) for the vaporiation of one mole of water at 1atm and 373K is zero. Reason: For all isothermal processes DeltaT=0

Calculate the work done when done when 1.0 mol of water at 373K vaporises against an atmosheric pressure of 1.0atm . Assume ideal gas behaviour.

For a cyclic process,the condition is (A) Delta U=0 only (B) Delta H=0 only (C) Delta U>0 and Delta H>0 (D) both Delta U=0 and Delta H=0

In which of the following cases,the reaction is spontaneous at all temperature [A] Delta H>0,Delta S>0 ; [B] Delta H 0 ; [C] Delta H>0,Delta S<0