Let 1.00 kg of liquid water at 100∘C be converted to steam at 100∘C 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 1.00×10−3 m3 as a liquid to 1.671 m3 as steam.
1. How much work is done by the system during this process?
Let 1.00 kg of liquid water at 100∘C be converted to steam at 100∘C 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 1.00×10−3 m3 as a liquid to 1.671 m3 as steam.
1. How much work is done by the system during this process?
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=1×10−3m3
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(Vf−Vi)
=1.01×105(1.670−0.001)
=169kJ.
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=1×10−3m3
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(Vf−Vi)
=1.01×105(1.670−0.001)
=169kJ.
