Question

this is my 2020 1st question...xD
Hey Friends, plz Answer mY question ❤

➡ Calculate the work done when 1.0 mole of water at 373 J vaporizes against an atmospheric pressure of 1.0 atmosphere. Assume ideal gas behaviour. ​

Answer 1

1 Mole of water = 18 g of water

Volume of water = 18g/1g or Ml

V1 = 18 ml

V2 = NRT /p

1.0×0.0821×373/ 1.0 = 30.6 litre

W = p ext × delta v = - (1.0) (30.6)litre - ATM

= (-30.6)×101.3k = -3098.3k

Answer 2

Answer:

Volume of 1 mol of water at 373 K, 1 atm (V1) = 18cm3 = 0.018 L

Volume of 1 mol of water in vapour state at 373 K,

1 = RTP=0.0821cm3atmK−1mol−1×373K1

atm=30.62 L

Now w=−P(V2−V1) = 1(30.62 - 0.018) = -30.6 L-atm

= -30.6 × 101.3 = -3100 J (Hydrated salt)