A large research balloon containing 2.00 X 103 m3 of helium gas at 1.00 atm and

Question

A large research balloon containing 2.00 X 103 m3 of helium gas at 1.00 atm and a temperature of 15.0oC rises rapidly from ground level to an altitude at which the atmospheric pressure is only 0.900 atm. Assume the helium behaves like an ideal gas and the balloon’s ascent is too rapid to permit much heat exchange with the surrounding air.
Calculate the volume and the temperature of the gas at the higher altitude.

I think I correctly figured out the volume, but the I don't think i got the right answer for the temperature. I may not be using the right form of the equation

Explanation / Answer

The process is adiabiatic P_1 V1 ^ = P_2 V2 ^      For helium = 1.67 , P_1 = 1 atm P _2 = 0.99 atm      V_1 = 2.00*10^3 m^3 a ) Volume   V2 ^   = V1 ^ ( P_1 / P_2 )                                   = V_1 * ( P_1 / P_2 )^ 1/          = 2.00*10^3 m^3 ( 1.00/ 0.99) ^ 1/ 1.67         = 2.13*10^3 m^3 b ) Temperature is          T_2 = T_1 ( P_2 / P_1 ) * ( V_2 / V_1 )                = 288.15 K * ( 0.99 atm / 1.00atm ) * ( 2.13 *10^3 m^3 / 2.00*10^3 m^3 )               = 303.8K                                 

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