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)`.

1g of water at 100^(@)C is converted into steam at the same temperature. If the volume of steam is 1671 cm^(3) , find the change in the internal energy of the system. Latent of steam =2256 Jg^(-1) . Given 1 atmospheric pressure = 1.013xx10^(5) Nm^(-2)

When 1 g of water at 100^@ C gets converted into steam at the same temperature, the change in volume is approximately

1 g of water (volume 1 cm^(3) becomes 1671 cm^(3) of steam when boiled at a pressure of 1atm. The latent heat of vaporisation is 540 cal/g, then the external work done is (1 atm = 1.013 xx 10^(5) N/ m^(2))

2kg of water is converted into steam by bolling at atmosperic pressure. The volume changes from 2xx10^(-3) m^(-3) to 3.34 m^(3) . The work done by the system is about

1.0 m^(3) of water is converted into 1671 m^(3) of steam at atmospheric pressure and 100^(@)C temperature. The latent heat of vaporisation of water is 2.3xx10^(6) J kg^(-1) . If 2.0 kg of water be converted into steam at atmospheric pressure and 100^(@)C temperature, then how much will be the increases in its internal energy? Density of water 1.0xx10^(3) kg m^(-3) , atmospheric pressure = 1.01xx10^(5) Nm^(-2) .

The volume of steam produced by 1g of water at 100^(@)C is 1650 cm^(3) . Calculate the change in internal energy during the change of state. Given J= 4.2xx10^(7) erg . cal.^(-1), d= 981 cm^(s-2) . Latent heat of stream = 540 cal. G^(-1)

Calculate the entropy change for vapourisation of water if latent heat of vapourisation for water is 540 cal/g. ( K_(b) for water =0.51 K. kg "mole"^(-1) )

Calculate the entropy change when one mole of water at 373 K is converted into steam. Latent heat of vaporisation of water (DeltaH_(v)) is 40.7 xx 10^(3) J mol^(-1)