What is U when 2-0 mole of liquid water vaporises at 100- The heat of vaporisation- H vap- of water at 100- C is 40-66 kJ mol-1

H_2O (l) nbsp; → H_2O (g) : n_g = 1 for 2 moles 2H_2O (l) nbsp; → 2H_2O (g) : n_g = 2 U = H n_g × RT = 2 × 40.66 2 × 8.314 × 373 × 10^ 3 = 75. ...

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Standard XI

Chemistry

Enthalpy

What is U whe...

Question

What is $△ U$ when 2.0 mole of liquid water vaporises at $100^{\circ} ?$ The heat of vaporisation, $△ H$ vap. of water at $100^{\circ} C$ is $40.66$ $k J m o l^{- 1}$

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75.11

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Solution

The correct option is B 75.11
$H_{2} O (l) \to H_{2} O (g) : △ n_{g} = 1$
for 2 moles
$2 H_{2} O (l) \to 2 H_{2} O (g) : △ n_{g} = 2$
$△ U = △ H - △ n_{g} \times R T = 2 \times 40.66 - 2 \times 8.314 \times 373 \times 10^{- 3} = 75.11 k J$

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Similar questions

Q. What is

△ U

when 2.0 mole of liquid water vaporises at

100^{\circ} ?

The heat of vaporisation,

△ H

vap. of water at

100^{\circ} C

40.66

k J m o l^{- 1}

Q. What is

△ U

when 2.0 mole of liquid water vaporises at

100^{\circ} ?

The heat of vaporisation,

△ H

vap. of water at

100^{\circ} C

40.66

k J m o l^{- 1}

Q. Find the value of

Δ E

when 2.0 mole of liquid water vaporises at

100^{o} C

is:

Given: The heat of vaporisation

(Δ H v a p .)

of water at

100^{o} C

40.66 k J m o l^{- 1}

Q. What is

Δ U

when

2.0

mole of liquid water vaporises at

100^{o} C

? The heat of vaporisation

(Δ H_{v a p})

of water at

100^{o} C

40.66 K J {m o l}^{- 1}

Q. What is

Δ E

when

2.0

mole of liquid water vaporises at

100^{o} C

? The heat of vaporisation, (

Δ H

) of water at

100^{o} C

40.66 k J {m o l}^{- 1}

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Enthalpy

Standard XI Chemistry

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What is △U when 2.0 mole of liquid water vaporises at 100∘? The heat of vaporisation, △H vap. of water at 100∘C is 40.66 kJ mol−1

The correct option is B 75.11 H2O(l)→H2O(g) :△ng=1 for 2 moles 2H2O(l)→2H2O(g) :△ng=2 △U=△H−△ng×RT=2×40.66−2×8.314×373×10−3=75.11 kJ

What is $△ U$ when 2.0 mole of liquid water vaporises at $100^{\circ} ?$ The heat of vaporisation, $△ H$ vap. of water at $100^{\circ} C$ is $40.66$ $k J m o l^{- 1}$

The correct option is B 75.11
$H_{2} O (l) \to H_{2} O (g) : △ n_{g} = 1$
for 2 moles
$2 H_{2} O (l) \to 2 H_{2} O (g) : △ n_{g} = 2$
$△ U = △ H - △ n_{g} \times R T = 2 \times 40.66 - 2 \times 8.314 \times 373 \times 10^{- 3} = 75.11 k J$