Calculate the final temperature of 12.0 g of Argon (considered as an ideal gas) that is expanded reversibly and adiabatically from V1 = 1.00 dm3 at T1 = 273.15 K to V2 = 3.00 dm3 . CV, m(Ar) = (3/2) * R.

Explanation:

For adiabatic change the expression is  

=

for Argon γ = 1.67

273.15 x 1 = T₂ x

T₂ = 273.15 /

= 273.15/ 2.0877

= 130.83 K.

catya that's not even a b or c

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