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Standard XII

Chemistry

Introduction

The value for...

Question

The value for $△$ U for the reversible isothermal evaporation of 90 g water at $100^{0} C$ will be : ( $△ H_{e v a p}$ of water = 40.8 kJ $m o l^{- 1}, R = 8.314 J K^{- 1} m o l^{- 1}$ ).

4800kJ

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188.494 kJ

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40.8 kJ

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125.03 kJ

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Solution

The correct option is B 188.494 kJ
$Δ H$ for 18 g water = 40.8kJ
for 90 g water = $\frac{40.8}{18} \times 90$ = 204 kJ
n for 90 g water = 90/18 =5
$Δ n_{g}$ = 5-0=5
$δ H = Δ U + Δ n_{g} R T$ = 188.494 kJ

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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}

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

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:

Q. The enthalpy of vaporization of water at

100^{0} C

is 40.63 KJ

m o l^{- 1}

. The value

Δ U

for this process would be:

Q. Calculate w and

△ U

for the conversion of 1 mol of water at

100^{0} C

to steam at 1 atm pressure. Heat of vaporization of water at

100^{0} C

is 40.670 kJ

{m o l}^{- 1}

. Assume ideal gas behavior.

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The value for △U for the reversible isothermal evaporation of 90 g water at 1000C will be : (△Hevap of water = 40.8 kJ mol−1,R=8.314JK−1mol−1).

The correct option is B 188.494 kJΔH for 18 g water = 40.8kJfor 90 g water = 40.818×90 = 204 kJn for 90 g water = 90/18 =5Δng = 5-0=5δH=ΔU+ΔngRT= 188.494 kJ

The value for $△$ U for the reversible isothermal evaporation of 90 g water at $100^{0} C$ will be : ( $△ H_{e v a p}$ of water = 40.8 kJ $m o l^{- 1}, R = 8.314 J K^{- 1} m o l^{- 1}$ ).

The correct option is B 188.494 kJ
$Δ H$ for 18 g water = 40.8kJ
for 90 g water = $\frac{40.8}{18} \times 90$ = 204 kJ
n for 90 g water = 90/18 =5
$Δ n_{g}$ = 5-0=5
$δ H = Δ U + Δ n_{g} R T$ = 188.494 kJ