What is the value of the change in internal energy at 1 atm in the following process-H2Ol-323 K - H2Og-423 KGiven - Cv- mH2O-1 -75-0 JK-1- Cp-mH2O- g -33-314 JK-1 mol-1- - Hvapat 373 K - 40-7 kJ-mol

We can break it down into three parts. nbsp; (i) H_2O (l, 323 K) → H_2O(l, 373 K) (ii) nbsp;H_2O (l, 373 K) → H_2O(g, 373 K) (iii) nbsp;H_2O (g, 373 K) ...

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Byju's Answer

Standard XI

Chemistry

Heat Capacity at Constant Volume

What is the v...

Question

What is the value of the change in internal energy at 1 atm in the following process?
$H_{2} O (l, 323 K) \to H_{2} O (g, 423 K)$
Given : $C_{v, m} (H_{2} O, 1) = 75.0 J K^{- 1}; C_{p, m} (H_{2} O, g) = 33.314 J K^{- 1} m o l^{- 1}; Δ H_{v a p} at 373 K = 40.7 k J / m o l$

42.910 kJ/mol

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43.086 kJ/mol

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42.600 kJ/mol

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49.600 kJ/mol

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Solution

The correct option is C $42.600 kJ/mol$
We can break it down into three parts.
(i) $H_{2} O (l, 323 K) \to H_{2} O (l, 373 K)$
(ii) $H_{2} O (l, 373 K) \to H_{2} O (g, 373 K)$
(iii) $H_{2} O (g, 373 K) \to H_{2} O (g, 423 K)$
now,
in(i),
$q = n C_{v} △ T = 1 \times 75 \times 50 = 3750 J = 3.75 k J$
$w :$ negligible.
$∴ △ E = 3.75 k J$
in (ii),
$q = m L = △ H_{v a p} = 40.7 k J$
$w = - P_{e x t} △ V = - 10^{5} P a (22.4 \times \frac{373}{273} \times 10^{- 3} m^{3})$
$= - 3.06 k J$
$△ E = + 37.64 k J$

in (iii)
$q = n C_{p} △ T = 1 \times 33.314 \times 50$
$= 1.666 k J$
$ω = - 10^{5} P a (22.4 \times \frac{423}{273} - 22.4 \times \frac{373}{273}) \times 10^{- 3}$
$= - 0.410 k J$
$△ E = + 1. 256 k J$
$△ E_{t o t a l} = + 42.646 k J$

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Similar questions

Q. What is the value of the change in internal energy at 1 atm in the following process?

H_{2} O (l, 323 K) \to H_{2} O (g, 423 K)

Given :

C_{v, m} (H_{2} O, 1) = 75.0 J K^{- 1}; C_{p, m} (H_{2} O, g) = 33.314 J K^{- 1} m o l^{- 1}; Δ H_{v a p} at 373 K = 40.7 k J / m o l

Q. The molar enthalpy change for

H_{2} O (1) ⇌ H 2 O (g)

373

K and

1

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41

kJ/mol. Assuming ideal behavior, the internal energy change for vaporization of

1

mol of water at

373

K and

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atm in kJ

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?

Q. The molar enthalpy change for

H_{2} O (l) ⇌ H_{2} O (g)

373

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1

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41 k J / m o l

. Assuming ideal behavior, the internal energy change for vaporization of

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Q. Calculate the standard internal energy change for the reaction

C_{(g r a p h i t e)} + H_{2} O (g) \to C O (g) + H_{2} O (g)

Δ_{f}^{o}

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H_{2} O (g) = - 241.8 k J {m o l}^{- 1}

C O (g) = - 110.5 k J {m o l}^{- 1}

R = 8.314 J K^{- 1} {m o l}^{- 1}

Q. The molar heat capacity of oxygen gas is given by the expression

C_{v} = a + b T + c T^{2}

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What is the value of the change in internal energy at 1 atm in the following process? H2O(l,323 K)→H2O(g,423 K) Given : Cv,m(H2O,1)=75.0 JK−1;Cp,m(H2O,g)=33.314JK−1mol−1;ΔHvapat373K=40.7kJ/mol

What is the value of the change in internal energy at 1 atm in the following process?
$H_{2} O (l, 323 K) \to H_{2} O (g, 423 K)$
Given : $C_{v, m} (H_{2} O, 1) = 75.0 J K^{- 1}; C_{p, m} (H_{2} O, g) = 33.314 J K^{- 1} m o l^{- 1}; Δ H_{v a p} at 373 K = 40.7 k J / m o l$