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When 1 g of water 100^(@) C is converted into steam at100^(@)C,. It occupies a volumn of 1671 cc at normal atmospheric pressure . Find the increase in internal energy of the molecules of steam. |
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Answer» Solution :`1" atmosphere" = 1.013xx10^(5) Nm^(-2)` , volume of 1 gm of water, `V_(1) = 1cc = 10^(-6)m^(3)`, volume of steams = `1671 cc =1671xx10^(-6) m^(3)` External WORK done dW = `P(V_(2)-V_(1)) = 1.013xx10^(5)(1671xx10^(-6) -1xx10^(-6))=1.013xx167=169.2 J`. Specific latent heat of vaporisation of steam=540cal/g. So, heat supplied to CONVERT 1g of water into steam. `Delta Q = 540xx4.2J=2268J` By first LAW of THERMODYNAMICS `Delta U = Delta Q-Delta W = 2268-169.2=2098.8J` |
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