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 Â
  [tex]T_{final} = 278.5 \ K[/tex] Â
Explanation:
From the question we are told that
   The atmospheric pressure is [tex]P_a = 1.01 *10^{5} \ Pa[/tex]
    The temperature heated to [tex]T = 32.0^oC = 305 \ K[/tex] Â
    The new atmospheric pressure is  [tex]P_a__{n}} = 8.0 * 10^{4} \ Pa[/tex]
    The value of g is  [tex]g = 1.40[/tex]
    The rate of cooling is  [tex]r = \frac{1}{100} \ ^oC/m[/tex]
The adiabatic relation for this process is mathematically represented as
      [tex]P^{1- g}_{initial} T^{g}_{initial} = P^{1-g} _{final} T^{g}_{final}[/tex]
Now  [tex]P}_{initial} = P_ a[/tex] ,  [tex]T_{initial} = T[/tex],[tex]P _{final} = P_a__{n}}[/tex] ,  [tex]T_{final} = ?[/tex]
substituting value Â
      [tex](1.01*10^{5})^{1-1.40} * (305) ^{1.40} = (8.0*10^{4})^{1-1.40} * T_{final} ^{1.40}[/tex]
     [tex]T_{final} ^{1.40} = 2646.7[/tex]
 =>   [tex]T_{final} ^{\frac{1.40}{1.40} } = 2646.7^{(\frac{1}{1.40})}[/tex]    Â
    [tex]T_{final} = 278.5 \ K[/tex] Â
