On a warm summer day, a large mass of air (atmospheric pressure 1:01 105 Pa) is heated by the ground to 32.0 C and then begins to rise through the cooler surrounding air. (This can be treated approximately as an adiabatic process; why?) Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8:00 104 Pa. Assume that air is an ideal gas, with g = 1.40. (This rate of cooling for dry, rising air, corresponding to roughly 1 C per 100 m of altitude, is called the dry adiabatic lapse rate.)

The process is adiabatic because there no heat transfer into the system or out of the system in that there is only one system involved in this process that is air.

The temperature of the air mass when it has risen is  

Explanation:

From the question we are told that

      The atmospheric pressure is

       The temperature heated to  

       The new atmospheric pressure is  

       The value of g is  

        The rate of cooling is  

The adiabatic relation for this process is mathematically represented as

Now   ,   , ,  

substituting value  

 =>            

yes, water is wet, just as it makes other things wet, it is wet itself, kinda like ice, when ice is in water, the ice makes water cold, and ice is cold, therefore with that example, water is wet