Question

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.

Answer 1

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.