A) \[-1.54\,kJ\]
B) \[1.54\,kJ\]
C) \[1.25\,kJ\]
D) \[-1.35\,kJ\]
Correct Answer: A
Solution :
[a]volume of 0.5 mole of steam at I atm pressure \[=\frac{nRT}{P}=\frac{0.5\times 0.0821\times 373}{1.0}=15.3L\] |
Change in volume = vol. of steam - vol. of water =\[15..3-negligible=15.3L\] |
Work done by the system, |
\[w=P\times \]volume change |
\[=\overset{ext}{\mathop{1}}\,\times 15.3=15.3\]litre-atm |
=\[15.3\times 101.3J=1549.89J\] |
?w? should be negative as the work has been done by the system on the surroundings. |
= -1549.89 J = -1.54 kJ |
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