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Question

Let 1.00 kg of liquid water at 100C be converted to steam at 100C by boiling at standard atmospheric pressure (which is 1.00 atm or 1.01 × 105 Pa) in the arrangement as shown. The volume of that water changes from an initial value of 1.00×103 m3 as a liquid to 1.671 m3 as steam.

1. How much work is done by the system during this process?


A

100 kJ

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B

169 kJ

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C

4200 kJ

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D

4.2 kJ

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Solution

The correct option is B

169 kJ


The weight of the lead balls on the piston remains a constant though the process. Since the weight of these balls have to be supported by the pressure of the gas, the pressure of the system is also a constant.

Initial condition
Volumeinitial=Vi=1×103m3
Pressureinitial=pi=1.01×105Pa
Volumefinal=Vf=1.671m3
Pressurefinal=pf=1.01×105Pa
Work done by the gas =W=vfvipdV

W=vfvipdV(since p is constant)
=1.01×105(VfVi)
=1.01×105(1.6700.001)

=169kJ.


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