alclti0f1s 1ons Modern Approach 416 =1-013 x 10 dynes/cm2 C = 18 m. V=volume of 1 mole of steam at 100°C 373 7x 22400=30605 mL. Kvolume of 1 mole of water at 100°C 16 vol. 0 wot ad oo Chare Now we have, W=-p(V2- V) = -1-013 x 10° x (30605 - 18) ergs -1-013 x 10° x 30587 calories 4-18x 10 -741 calories. Again we have,
First of all for pressure units are converted to cms
Then volume of one mole H2O (1*2+16=18)ml is taken
Then for steam (which is in gaseous state) conversion are made
0 to 100 with...