NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice
Calculate the work done when 2.5 mol of H 2 O vaporizes at 1.0 atm and 25 ° C . Assume the volume of liquid H 2 O is negligible compared to that of vapour. Given 1 L atm = 101.3 J and R = 0.082 L atm m o l - 1 K - 1 .
Options
- A6190 kJ
- B6.19 kJ
- C61.1 kJ
- D5.66 kJ
Correct answer
B. 6.19 kJ
Step-by-step solution
Volume of water vapour at 1.0   atm and 298   K is given by V = n R T P = 2 .5 mol × 0 .082 L atm mol − 1 K − 1 × 298 K 1 atm = 61.09 L Now change volume Δ V = V f i n a l - V i n i t i a l = 61.09 L − 0 L = 61.09 L So work done against constant pressure of 1 atm = P Δ V = 1   atm   ×   61.09   L = 61.09   L   atm = 61.09   L   atm × 101.3   J 1   L   atm = 6.19 kJ