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205
NUMERICALS WITH SOLUTION
Three moles of an ideal gas are expanded Solution :
isothermally from a volume of 300 cm3 to
W = –pex (V2 – V1)
2.5 L at 300 K against a pressure of 1.9
36.50 J = –1.216 × 105 Nm–2
atm. Calculate the work done in L atm and
× (V2 – 500 × 10–6 m3)
joules.
= –1.216 × 105 Nm–2 × V2 +
Given :
1.216 × 105 Nm–2 × 500 ×
n = 3 moles,
10–6 m3
3
V1 = 300 cm = 0.300 L,
= –1.216 × 105 × V2 + 60.8
V2 = 2.5 L
∴ 1.216 × 105 V2 = 60.8 – 36.50
T = 300 K,
24.3
V2 = 1.216 × 10 5
pex = 1.9 atm.
To find :
= 200 × 10–6m3 = 200cm3
Work done (W) = ?
∴ V2 = 200 cm3
(in L and J)
Solution :
*3. Calculate the maximum work when
W = –pex (V2 – V 1)
24 g of oxygen are expanded isothermally
= –1.9 atm (2.5 L – 0.3 L)
and reversibly from a pressure of
= –1.9 × 2.2 L atm.
1.6 × 105 Pa to 100 kPa at 298 K.
W = –4.18 L atm
Given :
Q 1 L atm = 101.3 J
W O 2 = 24g,
∴ W = –4.18 × 101.3
P 1 = 1.6 × 105 Pa
W = –423.4J
= 1.6 × 102 kPa
= 160 kPa
*2. One mole of an ideal gas is compressed
3
=
100 kPa
P
2
from 500 cm against a constant pressure
T = 298 K.
of 1.216 × 105 Pa. The work involved in
the process is 36.50 J. Calculate the final To find :
Wmax = ?
volume.
Solution
:
Given :
WO 2 24
n = 1 mole,
no. of moles (n) =
=
= 0.75 mol
32
M
V1 = 500 cm3 = 500 × 10–6m3
P1
pex = 1.216 × 105 Pa (Nm–2),
Wmax = –2.303 × n × R × T × log10 P
2
W = 36.50 J.
= –2.303 × 0.75mol × 8.314 JK–1
To find :
160 kPa
V2 = ?
mol–1 × 298 K ×log
*1.
10
Chapter - 3 Chemical Thermodynamics And Energetics
100 kPa
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