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Arvind Chander
Asked: 2022-11-10T17:30:37+05:30 In:

Determine the value of `DeltaH` and `DeltaU` for the reversible isothermal evaporationof 90.0 g of water at `100^(@)C` . Assume that water vapour behave as an ideal gas and heat of evaporation of water is 540 cal `g^(-1) ` `( R = 2.0 cal mol^(-1)K^(-1))`

Determine the value of `DeltaH` and `DeltaU` for the reversible isothermal evaporationof 90.0 g of water at `100^(@)C` . Assume that water vapour behave as an ideal gas and heat of evaporation of water is 540 cal `g^(-1) ` `( R = 2.0 cal mol^(-1)K^(-1))`
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  1. Correct Answer – `Delta H = 48600 cal, Delta U = 44870cal`
    `DeltaH = 540 xx 90 cal = 48600 cal, DeltaU = DeltaH – Deltan_(g)RT`,
    `90 g H_(2) O = 5 ` moles , `Delta n_(g) = 5 ` for moles of `H_(2)O(l) rarr 5` moles of `H_(2)O(g)`

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Riddhi Naruka
Asked: 2022-10-30T16:37:22+05:30 In:

Determine the value of `DeltaH` and `DeltaU` for the reversible isothermal evaporation of `90.0g` of water at `100^(@)C`. Assume that water behaves as an ideal gas and heat of evaporation of water is `540 cal g^(-1) (R = 2.0 cal mol^(-1)K^(-1))`.

Determine the value of `DeltaH` and `DeltaU` for the reversible isothermal evaporation of `90.0g` of water at `100^(@)C`. Assume that water behaves as an ideal gas and heat of evaporation of water is `540 cal g^(-1) (R = 2.0 cal mol^(-1)K^(-1))`.
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  1. Total heat change
    `DeltaH = 90.0 xx 540 = 48600cal`
    Now, `DeltaH = DeltaU +P DeltaV`
    Here, `DeltaV =(V_(vapour) -V_(liquid)) = V_(vapour)`
    (Volume of liquid is negligible as compared to volume of vapor)
    `DeltaH = DeltaU +PV_(vapour) = DeltaU +nRT`
    `n = (90)/(18) = 5mol, R = 2 cal K^(-1) mol^(-1), T = 373 K`
    `:. DeltaH = DeltaU +5 xx2 xx 373`
    `DeltaH = DeltaU +3730`
    or `DeltaU = DeltaH – 3730 =48600 – 3730 = 44870cal`

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