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o-na [289]
3 years ago
12

Calculate the final temperature of 12.0 g of Argon (considered as an ideal gas) that is expanded reversibly and adiabatically fr

om V1 = 1.00 dm3 at T1 = 273.15 K to V2 = 3.00 dm3 . CV,m(Ar) = (3/2) * R.
Chemistry
1 answer:
Deffense [45]3 years ago
3 0

Answer:

Explanation:

For adiabatic change the expression is  

= PV^\gamma=constant

=(\frac{RT}{V})V^\gamma = constant

=TV^{\gamma-1}  = constant

=T_1V_1^{\gamma-1}  ==T_2V_2^{\gamma-1}

for Argon γ = 1.67

273.15 x 1 = T₂ x 3^{1.67-1}

T₂ = 273.15 / 3^{0.67}

= 273.15/ 2.0877

= 130.83 K.

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