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Transcript
206
= –2.303 × 0.75 × 8.314 × 298 ×
0.2041
Wmax = –873.4 J
*4.
Three moles of an ideal gas are
compressed isothermally and reversibly
to a volume of 2L. The work done is
2.983kJ at 22ºC. Calculate the initial
volume of the gas.
Given :
n = 3 mol,
V2 = 2L,
Wmax = 2.983 kJ = 2983 J
T = 22ºC + 273 = 295K
To find :
V1 = ?
Solution :
Wmax = –2.303 × n × R × T × log10
V2
V1
Given :
W N 2 = 2.8 × 10–2 kg
T = 300 K
P1 = 15.15 × 105 Nm–2
Wmax = – 17.33 kJ = –17330 J
To find :
P2 = ?
Solution :
W
2.8 × 10 –2
n=
=
=1
M
28 × 10 –3
Wmax = –2.303 × n × R × T × log10
–17330 J = –2.303 × 1mol × 8.314 JK –1mol –1
×300K× log10
∴
∴
*5.
17330J
= 5744.1 × log10
= 1.500
V1 = 1.5 × 2L = 3.0 L
15.15 × 10 5
P2
17330
5744.1
= log10
15.15 × 10 5
P2
3.0170
= log10
15.15 × 10 5
P2
Al (3.0170) =
= AL (0.1760)
15.15 × 10 5
P2
15.15 × 10 5
1.040 × 10 =
P2
3
2.8 × 10–2 kg of nitrogen is expanded
isothermally and reversibly at 300K from
15.15 × 105 Nm–2 when the work done
is found to be –17.33 kJ. Find the final
pressure.
Unique Solutions ®
15.15×10 5 Nm –2
P2
⎡
15.15×10 5 N ⎤
– ⎢ 2303× 1×8.314×300×log10
⎥
= ⎢
P2
⎣
⎦⎥
2983 J = –2.303 × 3 mol × 8.314 JK–1
2L
mol–1 × 295K × log10
V1
V
2983 = + 16945.2 × log10 1
2L
2983
V1
log10
=
16945.2
2L
= 0.1760
V1
2L
P1
P2
15.15 × 10 5
∴
P 2=
∴
= 14.5673 × 102 Nm–2≈ 1456.7 Nm–2
P2= 1456.7 Nm –2
1.040 × 10 3
S.Y.J.C. Science - Chemistry - Part I