Answer:
snow
Explanation:
Since the process undergoes adiabatic expansion, hence q = 0 and ΔU = w.
We can sole this problem using the following derivation:
[tex]ln(\frac{T_2}{T_1} )=-(\gamma -1)ln(\frac{V_f}{V_i} )=-(\gamma -1)ln(\frac{T_2}{T_1}\frac{P_i}{P_f} )\\=-(\gamma -1)ln(\frac{T_2}{T_1})-(\gamma -1)ln(\frac{P_i}{P_f})\\=-(\frac{\gamma -1}{\gamma})ln(\frac{P_i}{P_f})\\=-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\ln(\frac{T_2}{T_1} )==-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\Substituting\ values:\\\\[/tex]
[tex]ln(\frac{T_2}{T_1} )=-(\frac{\frac{28.86}{28.86-8.314} -1}{\frac{28.86}{28.86-8.314}})ln(\frac{0.802\ atm}{0.602\ atm})=-0.0826\\\\ln(\frac{T_2}{T_1} )=-0.0826\\\\Taking\ exponential\ of\ both \ sides:\\\\\frac{T_2}{T_1} =e^{-0.0826}\\\\T_2=0.9207T_1\\\\T_2=0.9207*288\\\\T_2=265\ K\\[/tex]
Since T2 = 265 K, we should expect a snow