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Standard XII

Chemistry

Properties of Enthalpy

For the rever...

Question

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

45870 c a l

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56870 c a l

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42000 c a l

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44870 c a l

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Solution

The correct option is D $44870 c a l$
Given,
$W a t e r ⟶ v a p o u r$

Moles before evaporation $\frac{90}{18} = 5$ 0

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

Thus, heat given at constant pressure:

$Δ H = h e a t o f e v a p o r a t i o n \times a m o u n t e v a p o r a t e d$

$= 540 \times 90$

$Δ H = 48600 c a l$

Also, $Δ H = Δ U + Δ n_{(g)} R T$

$Δ U = 48600 - Δ n_{(g)} R T$

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

$= 48600 - 3730$

$Δ U = 44870 c a l$

Suggest Corrections

Similar questions

Q. 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

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. 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

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Q. What is value of

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

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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. 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. For water

Δ_{v a p} H = 41 k J {mol}^{- 1}

373 K

and 1 bar pressure. Assuming that water vapour is an ideal gas the that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ___

k J {mol}^{- 1}

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[Use :

R = 8.3 J {mol}^{- 1} K^{- 1}

]

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For the reversible isothermal evaporation of 90.0 g of water at 100∘C. The value of ΔU for the reaction is:[Assuming that water vapours behave as an ideal gas and heat of evaporation of water is 540calg−1 and R=2calmol−1K−1]

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