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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.

Respuesta :

Answer:

Explanation:

For adiabatic change the expression is  

= [tex]PV^\gamma=constant[/tex]

[tex]=(\frac{RT}{V})V^\gamma = constant[/tex]

[tex]=TV^{\gamma-1} = constant[/tex]

[tex]=T_1V_1^{\gamma-1} =[/tex][tex]=T_2V_2^{\gamma-1}[/tex]

for Argon γ = 1.67

273.15 x 1 = T₂ x [tex]3^{1.67-1}[/tex]

T₂ = 273.15 / [tex]3^{0.67}[/tex]

= 273.15/ 2.0877

= 130.83 K.

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