Question

assume that water vapour behaves like as an ideal gas and heat of evaporation of water is 540 cal g^-1​

Answer 1

The complete question is:

What is the value of ΔE (heat change at constant volume) for reversible isothermal evaporation of 90g water at 100 ∘C Assuming water vapour behaves as an ideal gas and (ΔHvap) water = 540calg−1

Given:

90g  of water

The heat of vaporization = ΔHvap = 540calg−1

Temperature = 100 °C

To find:

What is the value of ΔE (heat change at constant volume) for reversible isothermal evaporation of 90g water at 100 ∘C.

Solution:

Weight of water = 90 g

The heat of vaporization = ΔHvap = 540calg−1 = 540 × 90

Temperature = 100 °C = 100 + 273 = 373 K

ΔU = ?

We use the formula,

ΔH = ΔU +ΔngRT

540 × 90 = ΔU + 5 × 2 × 373

⇒ 48600 = ΔU + 3730

⇒ ΔU = 48600 - 3730

∴ ΔU = 44870 cal

The heat change at constant volume) for reversible isothermal evaporation of 90g water at 100 ∘C  is 44870 cal