If water vapour is assumed to be a perfect gas- molar enthalpy changes for vaporization of 1 - of water at 1 - - 100- - 41 - -1- Calculate the internal energy change- when 1 - of water is vaporized at 1 - pressure and 100 -

Name: NCERT - Grade 11 - Chemistry - Thermodynamics - Q5
Uploaded: 2022-07-04T10:36:42+05:30
Description: Internal energy change The given change, 𝐻_2𝑂 (𝑙) → 𝐻_2𝑂(𝑔) nbsp;𝑛(𝑔)=1 − 0=1 𝑚𝑜𝑙 ∵ 𝐻= 𝑈 + 𝑛_𝑔 𝑅𝑇 or U = 𝐻 − 𝑛_𝑔 RT Given, 𝐻=41 𝑘𝐽 𝑚𝑜𝑙^−1, 𝑇=100 ^∘𝐶 ...

Internal energy change The given change, 𝐻_2𝑂 (𝑙) → 𝐻_2𝑂(𝑔) nbsp;𝑛(𝑔)=1 − 0=1 𝑚𝑜𝑙 ∵ 𝐻= 𝑈 + 𝑛_𝑔 𝑅𝑇 or U = 𝐻 − 𝑛_𝑔 RT Given, 𝐻=41 𝑘𝐽 𝑚𝑜𝑙^−1, 𝑇=100 ^∘𝐶 ...

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Byju's Answer

Standard VI

Geography

Isotherms

If water vapo...

Question

If water vapour is assumed to be a perfect gas, molar enthalpy changes for vaporization of $1 m o l$ of water at $1 b a r$ 𝑎𝑛𝑑 $100^{\circ} C$ 𝑖𝑠 $41 k J m o l^{- 1} .$ Calculate the internal energy change, when $1 m o l$ of water is vaporized at $1 b a r$ pressure and $100^{\circ} C .$

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Solution

Internal energy change

The given change, $H_{2} O (l) \to H_{2} O (g)$

$△ n (g) = 1 - 0 = 1 m o l$

$∵ △ H = △ U + △ n_{g} R T$

or $△ U = △ H - △ n_{g} R T$

Given,

$△ H = 41 k J m o l^{- 1},$

$T = 100^{\circ} C = 373 K$

$∴ △ U = 41 (k J m o l^{- 1}) - 1 (m o l) \times 8.314 \times 10^{- 3} (k J m o l^{- 1} K^{- 1} ) \times 373 K$

$△ U = 37.898 k J m o l^{- 1}$

Final answer: $△ U = 37.898 k J m o l^{- 1}$

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