Engineering Chemistry S Datta

96 ENGINEERING CHEMISTRY
∆H 150°
= ∆H 100°
– 0.55 × 50
= 538 – 0.55 × 50 = 510.5 cal g –1 .
Example 11. Calculate the maximum work done for the isothermal reversible expansion
of one mole of an ideal gas at 25°C from 2 atm to 1 atm.
Sol. Maximum work = 2.303 RT ln P 1
P2
= 2.303 × 8.31 × 298 log 2 1
= 1.66 × 10 3 J = 1.66 kJ.
Example 12. Calculate the change of molar entropy during melting of ice. Given latent
heat – 19.14 J g –1 .
Sol. dS = q rev
T
=
19.14 × 18
273
= 1.26 J mol –1 .
Example 13. Calculate the change of molar entropy during the conversion of liquid
oxygen to oxygen gas at its b.p. (– 182.9°C). Given L = 12.19 J g –1 .
Sol. dS = q rev
T
12.19 × 32
=
=
(273182.9) −
12.19 × 32
90.1
= 4.33 J mol –1 .
Example 14. Calculate change of entropy in isothermal reversible expansion of one mole
of an ideal gas.
Sol. We know that ∆E = 0
Therefore, q = W = RT ln V 2
V1
dS = q rev
T
= R ln V 2
V1
[Positive quantity]
Example 15. At what temperature will water boil when the atmospheric pressure is
528 mm Hg? Given latent heat of vapourisation of water is 545.5 cal g –1 .
Sol. Here,
P 1
= 528 mm of Hg; P 2
= 1 atm = 760 mm of Hg.
T 1
= ?; T 2
= 100°C = 373 K; H v
= 545.5 cal g –1
= 545.5 × 18 cal mol –1
= 9819 cal mol –1 .
R = 1.987 cal K –1 mol –1 .
From the integrated form of Clausius-Clapeyron equation, we have
log P 2
P1
=
∆Hv
2.303R
L
N
M
1
−
T
1
T
1 2
O
Q
P