Question

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.)

Answer 1

Answer:

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  

   $T_{final} = 278.5 \ K$  

Explanation:

From the question we are told that

      The atmospheric pressure is $P_a = 1.01 *10^{5} \ Pa$

       The temperature heated to $T = 32.0^oC = 305 \ K$  

       The new atmospheric pressure is  $P_a__{n}} = 8.0 * 10^{4} \ Pa$

       The value of g is  $g = 1.40$

        The rate of cooling is  $r = \frac{1}{100} \ ^oC/m$

The adiabatic relation for this process is mathematically represented as

            $P^{1- g}_{initial} T^{g}_{initial} = P^{1-g} _{final} T^{g}_{final}$

Now   $P}_{initial} = P_ a$ ,   $T_{initial} = T$,$P _{final} = P_a__{n}}$ ,  $T_{final} = ?$

substituting value  

           $(1.01*10^{5})^{1-1.40} * (305) ^{1.40} = (8.0*10^{4})^{1-1.40} * T_{final} ^{1.40}$

          $T_{final} ^{1.40} = 2646.7$

 =>    $T_{final} ^{\frac{1.40}{1.40} } = 2646.7^{(\frac{1}{1.40})}$        

        $T_{final} = 278.5 \ K$