Home
Class 11
CHEMISTRY
A swimmer coming out from a pool is cove...

A swimmer coming out from a pool is covered with a film of water weighing about 18 g. how much heat must be supplied to evaporate this water at 298 K ? Calculate the internal energy of vaperization at `100^(@)C`. `Delta_(vap)H^(Theta)` for water at 373 K = 40.66 kJ `mol^(-1)`

Text Solution

Verified by Experts

The process of evaporation is `: 18 H_(2) O(l ) rarr 18 g H_(2)O(g) `
No. of moles of 18 g `H_(2) O = (18g )/(18 g mol^(-1))= 1 mol`
`Delta n_(g) = 1-0 =1 mol`
`:. Delta _(vap) U^(@) = V_(vap) H^(@) - Delta n_(g) RT = 40.66 k J mol^(-1) - (1 mol ) (8.314 xx 10^(-3) k J K^(-1) mol^(-1) ) ( 298 K) `
`= 40.66 k J mol^(-1) - 3.10 kJ mol^(-1) = 37. 56 kJ mol^(-1)`
Promotional Banner

Topper's Solved these Questions

  • THERMODYNAMICS

    PRADEEP|Exercise Problem 12|1 Videos
  • THERMODYNAMICS

    PRADEEP|Exercise Sample Problem 1|6 Videos
  • THERMODYNAMICS

    PRADEEP|Exercise Problem 10|1 Videos
  • STRUCTURE OF ATOM

    PRADEEP|Exercise Competition Focus (JEE (Main and Advanced)/Medical Entrance (IX. Assertion And Reason Type Questions (Type II))|12 Videos

Similar Questions

Explore conceptually related problems

A swimmer coming out from a pool is covered with a film of water weighing about 18g. How much heat must be supplied to evaporate this water at 298 K ? Calculate the internal energy of vaporisation at 298K. Delta_("vap")H^(Ө) for water at 298K =44.01 kJ "mol"^(-1)

A swimmer coming out from a pool is covered with a film of water weighing about 80g . How much heat must be supplied to evaporate this waateer ? If latent heat of evaporation for H_(2)O is 40.79 kJ mol^(-1) at 100^(@)C .

18g of water is takento prepare the tea. Find out the internal energy of vaporisation at 100^(@) C. (Delta_(vap)H for water at 373 K 373 K is 40.66kJ mol^(-1))

Calculate the entropy change of n -hexae when 1mol of it evaporates at 341.7K (Delta_(vap)H^(Theta) = 290.0 kJ mol^(-1))

Calculate the entropy change for vaporization of 1mol of liquid water to stem at 100^(@)C , if Delta_(V)H = 40.8 kJ mol^(-1) .

The internal energy change (in J) when 90 g of water undergoes complete evaporation at 100^(@)C is ______ . (Given : Delta H_("vap") for water at 373 K = 41 kJ/ mol , R = 8.314 JK^(-1) mol^(-1) )

What is Delta U when 2.0 mole of liquid water vaporises at 100^(@)C ? The heat of vaporisation (Delta H_("vap".)) of water at 100^(@)C is 40.66 KJmol^(-1) .

Knowledge Check

  • 18 g of water is taken to prepare, the tea. Find out the internal energy of vaporisation at 100^(@)C. (Delta_(vap)H^(ө) for water at 373K=40.66kJmol^(-1))

    A
    `37.56 kJ` `mol^(-1)`
    B
    `-37.56" kJ "mol^(-1)`
    C
    `43.76" kJ "mol^(-1)`
    D
    `-43.76" kJ "mol^(-1)`
  • 18g of water is takento prepare the tea. Find out the internal energy of vaporisation at 100^(@) C. (Delta_(vap)H for water at 373 K 373 K is 40.66kJ mol^(-1))

    A
    `37.56kJmol^(-1)`
    B
    `-37.56kJmol^(-1)`
    C
    `43.76kJmol^(-1)`
    D
    `3-43.76kJmol^(-1)`
  • A boy after swimming comes out from a pool covered with a film of water weighing 80 g .How much heat must be supplied to evaporate this water ? (Delta_(v) H^(@)=40.79 kJmol^(-1))

    A
    `1.61 xx10^(2) kJ`
    B
    `1.71 xx 10^(2)kJ`
    C
    ` 1.81xx10^(2) kJ `
    D
    ` 1.91 xx 10^(2) kJ `