The values of $Δ H$ and $Δ U$ for the reversible isothermal evaporation of 90.0 g of water at $100^{o}$ C are $x$ $c a l$ & $y$ $c a l$ . Assume that water vapour behaves as an ideal gas and heat of evaporation of water is 540 cal $g^{- 1}$ . $(R = 2 c a l {m o l}^{- 1} K^{- 1})$ . The value of $x + y$ is:

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Solution

$W a t e r ⇌ V a p o u r$

Mole before evaporation $\frac{90}{18} = 5$

The evaporation of 5 moles of water takes place reversible and isothermally into vapours.

Thus, heat given at constant pressure $(Δ H)$ = heat of evaporation x amount evaporated.

$Δ H = 540 \times 90$

$Δ H = 48600 c a l$

Also, $Δ H = Δ U + Δ n R T$

$Δ U = 48600 - Δ n R T$

$= 48600 - (5 \times 2 \times 373)$

$= 48600 - 3730$

$Δ U = 44870 c a l$

Therefore, $Δ H + Δ U = 93470$

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Similar questions

Q. For the reversible isothermal evaporation of 90.0 g of water at

100^{\circ} C

. The value of

Δ U

for the reaction is:

[Assuming that water vapours behave as an ideal gas and heat of evaporation of water is

540 c a l g^{- 1}

and

R = 2 c a l m o l^{- 1} K^{- 1}

]

Q. What is value of

△ U

(heat change at constant volume) for reversible isothermal evaporation of 90 g water at

100^{o} C .

Assuming water vapour behaves as an ideal gas and

(△ H_{v a p})_{w a t e r} = 540 c a l g^{- 1}

Determine the value of ∆H and ∆U for the reversible isothermal evaporation of 90.0g of water at 100°C. Assume that water vapour behave as am ideal gas and heat of eveporation of water is 540cal.

MY PROBLEM

Only tell me How ∆H is calculated here, because if I know∆H then I can use the equation

∆H = ∆U + nRT

In my book it is given that

∆H = 90.0 x 540

= 48600 cal

But I want to know that why they have multipled the mass to heat of eveporation to find out ∆H

Q. What is value of

△ U

(heat change at constant volume) for reversible isothermal evaporation of 90 g water at

100^{o} C .

Assuming water vapour behaves as an ideal gas and

△ H_{v a p})_{w a t e r} = 540 c a l g^{- 1}

Q. The latent heat of vapourization of water at

100^{o} C

540 c a l . g^{- 1}

. Calculate the entropy increase when one mole of water at

100^{o} C

is evaporated.

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The values of ΔH and ΔU for the reversible isothermal evaporation of 90.0 g of water at 100oC are x cal & y cal. Assume that water vapour behaves as an ideal gas and heat of evaporation of water is 540 cal g−1. (R=2cal mol−1K−1). The value of x+y is: