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Boiling water: Suppose 1.0 g of water va...

Boiling water: Suppose 1.0 g of water vaporizes isobarically at atmospheric pressure `(1.01xx10^5 Pa)`. Its volume in the liquid state is `V_i=V_(liqu i d)=1.0cm^3` and its volume in vapour state is `V_f=V_(vapour)=1671 cm^3`. Find the work done in the expansion and the change in internal energy of the system. Ignore any mixing of the stream and the surrounding air. Take latent heat of vaporization `L_v=2.26xx10^6J//kg`.

Text Solution

Verified by Experts

The correct Answer is:
A, B

Because the expansion takes place at constant pressure, the work done is
`W=int_(V_i)^(V_f)p_0dV=p_0int_(V_i)^(V^f)dV=p_0(V_f-V_i)`
`=(1.01xx10^5)(1671xx10^-6-1.0xx10^-6)`
`=169J`
`Q=mL_v=(1.0xx10^-3)(2.26xx10^6)`
`=2260J`
Hence, from the first law, the change in internal energy
`DeltaU=Q-W=2260-169`
`=2091J`
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Knowledge Check

  • 1 g of water, of volume 1 cm^3 at 100^@C , is converted into steam at same temperature under normal atmospheric pressure (= 1 xx 10^5 Pa) . The volume of steam formed equals 1671 cm^3 . If the specific latent heat of vaporisation of water is 2256 J/g, the change in internal energy is

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