One kg of water at `373K` is converted into steam at the same temperature. The volume `1 cm^(3)` of water becomes `1671 cm^(3)` on boiling. Calculate the change in internal energy of the system , if heat of vaporisation is `540 cal g^(-1)`. Given standard atmospheric pressure `=1.013xx10^(5) Nm^(-2)`.

One kg of water at `373K` is converted into steam at the same temperat

1.	One kg of water at `373K` is converted into steam at the same temperature. The volume `1 cm^(3)` of water becomes `1671 cm^(3)` on boiling. Calculate the change in internal energy of the system , if heat of vaporisation is `540 cal g^(-1)`. Given standard atmospheric pressure `=1.013xx10^(5) Nm^(-2)`.
Answer» Volume of `1kg` of water = `1000cm^(3) = 10^(-3) m^(3)` , Volume of `1kg` of steam = `10^(3) xx 1671 cm^(3) = 1.671 m^(3)` Change in volume , ` DeltaV = (1.671-10^(-3))m^(3)= 1.670 m^(3)`, Pressure , `P = 1"atm". = 1.01 xx 10^(5) Nm^(-2)` In expansion work done , `W= P Delta V = 1.01 xx10^(5) xx 1.67J = (1.686xx15^(5))/(4.2) cal = 4.015xx10^(4) cal` But `DeltaU = Q-W`(first law of thermodynamics) or `DeltaU = (5.4xx 10^(5) - 0.04015 xx 10^(5)) cal = 4.9985 xx 10^(5) cal`

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