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Transcript
```206
= â2.303 Ã 0.75 Ã 8.314 Ã 298 Ã
0.2041
Wmax = â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 22ÂºC. Calculate the initial
volume of the gas.
Given :
n = 3 mol,
V2 = 2L,
Wmax = 2.983 kJ = 2983 J
T = 22ÂºC + 273 = 295K
To find :
V1 = ?
Solution :
Wmax = â2.303 Ã n Ã R Ã T Ã log10
V2
V1
Given :
W N 2 = 2.8 Ã 10â2 kg
T = 300 K
P1 = 15.15 Ã 105 Nmâ2
Wmax = â 17.33 kJ = â17330 J
To find :
P2 = ?
Solution :
W
2.8 Ã 10 â2
n=
=
=1
M
28 Ã 10 â3
Wmax = â2.303 Ã n Ã R Ã T Ã log10
â17330 J = â2.303 Ã 1mol Ã 8.314 JK â1mol â1
Ã300KÃ log10
â´
â´
*5.
17330J
= 5744.1 Ã log10
= 1.500
V1 = 1.5 Ã 2L = 3.0 L
15.15 Ã 10 5
P2
17330
5744.1
= log10
15.15 Ã 10 5
P2
3.0170
= log10
15.15 Ã 10 5
P2
Al (3.0170) =
= AL (0.1760)
15.15 Ã 10 5
P2
15.15 Ã 10 5
1.040 Ã 10 =
P2
3
2.8 Ã 10â2 kg of nitrogen is expanded
isothermally and reversibly at 300K from
15.15 Ã 105 Nmâ2 when the work done
is found to be â17.33 kJ. Find the final
pressure.
Unique Solutions Â®
15.15Ã10 5 Nm â2
P2
â¡
15.15Ã10 5 N â¤
â â¢ 2303Ã 1Ã8.314Ã300Ãlog10
â¥
= â¢
P2
â£
â¦â¥
2983 J = â2.303 Ã 3 mol Ã 8.314 JKâ1
2L
molâ1 Ã 295K Ã log10
V1
V
2983 = + 16945.2 Ã log10 1
2L
2983
V1
log10
=
16945.2
2L
= 0.1760
V1
2L
P1
P2
15.15 Ã 10 5
â´
P 2=
â´
= 14.5673 Ã 102 Nmâ2â 1456.7 Nmâ2
P2= 1456.7 Nm â2
1.040 Ã 10 3
S.Y.J.C. Science - Chemistry - Part I
```
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