Question

If water vapour is assumed to be a perfect gas, molar enthalpy changes for evaporation of $ 1mol $ of water at 1 bar and $ {100^ \circ }C $ is $ 41kJ\,mo{l^{ - 1}} $ . Calculate the internal energy change when :
(1) $ 1mol $ of water is evaporated at $ 1bar $ pressure and $ {100^ \circ }C $ .
(2) $ 1mol $ of water is converted into ice.

Answer 1

Hint: In order to the question, we have to find the relation between the change in enthalpy and the change in internal energy. Universal Gas Constant is denoted by $ R $ .

Complete step by step solution:
Given values according to the question:
 $ n = 1mole $ (given moles of water)
Pressure = $ 1bar $
Temperature, $ T $ = $ {100^ \circ }C $
  $ T = 100 + 273K $
 $ T = 373K $
And, we have also the molar enthalpy change for vapourisation :
Change in enthalpy, $ \Delta H = 41kJ\,mo{l^{ - 1}} $
Now,
(1) we have to calculate the change in internal energy when $ 1mol $ of water is evaporated at $ 1bar $ pressure and $ {100^ \circ }C $ or $ \Delta U $ .
So, we have the formulae of Change in Enthalpy
 $ \Delta H = \Delta U + \Delta n(g)RT $
here, R is the Universal Gas Constant, which has the value of $ 8.314 \times {10^{ - 3}}kJ\,mo{l^{ - 1}} $ $ \therefore \Delta U = \Delta H - \Delta n(g)RT $
  $ = 41 - (1 \times 8.314 \times {10^{ - 3}} \times 373) $
   $ = 37.904kJ\,mo{l^{ - 1}} $
Hence, the value of change in the internal energy when $ 1mol $ of water is evaporated at $ 1bar $ pressure and $ {100^ \circ }C $ is $ 37.904kJ\,mo{l^{ - 1}} $ .
(2) In second part, we have to convert $ 1mol $ of water into ice:
when $ 1mol $ of water is converted into ice:
The change: $ {H_2}O(I) \to {H_2}O(s) $
Since during this process, there will be a negligible change in the volume, so, we can put:
 $ p\Delta V = \Delta nRT = 0 $
So, in this case, $ \Delta H = \Delta U $
 $ \therefore \Delta U = 41kJ\,mo{l^{ - 1}} $
Hence, the internal energy changes when $ 1mol $ of water is converted into ice.

Note:
Internal energy change is only proportional to the change in temperature under rather restricted circumstances. That being, the process (chemical reaction) must happen at constant volume. Then and only then does the proportionality work.