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CBSE
Class 11 Chemistry
Thermodynamics
Quiz 2
1
2
Standard enthalpy of vapourisation ΔH
vap
for water at 100oC is 40.66 kJmol
-1
. The internal energy of vapourisation of water at 100°C (in kJmol
-1
) is (Assume water vapour to behave like an ideal gas)
0%
43.76
0%
40.66
0%
37.56
0%
-43.76
Explanation
(c) 37.56
Explanation:
For gaseous reactants and products, we have a relation between standard enthalpy of vaporization (ΔH
vap
) and standard internal energy (ΔE) as-
ΔH
vap
= ΔE+ Δn
g
RT whereas,
Δn
g
= n
2
− n
1
, i.e., difference between no. of moles of reactant and product.
For vaporization of water,
H
2
O (l) → H
2
O(g)
Therefore, Δn
g
= 1 − 0 = 1
Therefore, ΔH
vap
= ΔE+ Δn
g
RT
⇒ ΔE = ΔH
vap
− Δn
g
RT = 40.63 − (1×8.314×10
-3
× 373) = 37.53 KJ/mol
Hence the value ΔE for this process will be 37.53KJ/mol.
The internal energy of vapourisation of water at 100°C (in kJmol
0%
43. 76
0%
40. 66
0%
37. 56
0%
-43. 76
Explanation
37. 56
0 h : 0 m : 1 s
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