Question

5 moles of helium and mixed with two moles of hydrogen to form a mixture take molar mass of helium m1 = 24 kg and that of helium and 2 equals to 2 kg the equivalent degree of freedom of the mixture is

Answer 1

The equivalent degree of freedom of the mixture is CV = 3.57

Explanation:

Mole fraction of Helium = 5 / 7 = 0.714

Mole fraction of Hydrogen = 2 / 7 = 0.285

Now (CV )mix) = 5/2 x 2R  + 3/2 x 5R / 7

(CV )mix) =  f (mix) / 2 R

(CV )mix) = 3.57 / 2 R

CV = 3.57

Thus the equivalent degree of freedom of the mixture is CV = 3.57

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Equilibrium concentration of hi, I2 and H2 is 0.7,0.1 and 0.1 moles/litre. Calculate the equilibrium constant for the reaction?

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Answer 2

Answer:24/7

Explanation:

Equivalent molar mass =   $\frac{∑nimi}{ni}$=  $\frac{5*4+2*2}{5+2}$ =  $\frac{24}{7}$ .