Engineering Chemistry S Datta

THERMODYNAMICS 97
∴ log 760
528 = 0.1582 = 9819
2.303 × 1.987
or, T 1
= 363 K.
L
N
M
1
T
1
1
−
373
Example 16. Calculate q, W, ∆E and ∆H for isothermal reversible expansion of an ideal
gas at 27°C from a volume of 10 dm 3 to 20 dm 3 against a constant external pressure of one
atmosphere?
Sol. (i) Since operation is isothermal and the gas is ideal so, ∆E = 0.
(ii) Now, from 1st law, ∆E = 0 or q = W.
But, for a reversible process, work done by system,
W = P(V 2
– V 1
)
= 1 atm (20 – 10) dm 3 = 10 atm dm 3 = 10 atm L
= 10 L atm ×
F
HG
O
Q
P
8.314 J
= 1012.7 J.
0.0821 L atmKJ (iii) q = W = 1012.7 J.
(iv) ∆H = ∆E + P∆V = 0 + 1012.7 J = 1012.7 J.
Example 17. A gas during expansion from 10 litre to 20 litre under 2 atmospheric
pressure absorbs 300 cal. heat energy. Find the change in internal energy.
Sol. Work done in an irreversible process is given by
W = P(V 2
– V 1
) = 2(20 – 10) = 20 lit.atm.
= (20 × 24.25) cal [Q 1 lit. atm = 24.25 cal]
= 485 cal.
From the 1st law of thermodynamics,
q = ∆E + W
or, ∆E = q – W = (300 – 485) cal = – 185 cal.
So, internal energy will decrease by an amount of 185 cal.
Example 18. The initial temperature and pressure for one mole of an ideal gas is (C v
=
5 cal) 27°C and 1 atm. Pressure is increased reversibly upto 7 atm and temperature is incremented
to 90°C. Calculate the entropy change for the process.
Sol. For one mole of an ideal gas the entropy change in a reversible process is
∆S = C p
ln T 2
T1
+ R ln P 1
P2
Here, T 1
= 27°C = 300 K; T 2
= 90°C = 363 K; P 1
= 1 atm; P 2
= 7 atm;
C p
= C v
+ R = (5 + 2) cal = 7 cal.
∴ ∆S = 7 ln 363
300 + 2 × ln 1 7 = – 2.5577 cal deg–1 .
Example 19. The value of K p
(equilibrium constant) of the reaction regarding ammonia
synthesis at 27°C is 3.49 × 10 –2 atm –1 . Thus find the value of K p
for the following reaction at
37°C.
1
2 N 2 (g) + 3 2 H 2 (g) NH 3
(g) + 11.2 Kcal.
Consider the reaction to be independent of heat and temperature.
I