a vessel contains a non linear triatomic gas. if 50% of gas disassociate into individual atom, then find new value of freedom by ignoring the vibrational mode and any further dissociation

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Let's assume that we have 1 mole of triatomic gas. Therefore 3Na (Avogardo's Number) is present initially. So, 0.5 moles has 1.5Na atoms. Now, 1 part of 0.5 moles remains undissociated. So. degree of dissociation = 0.5 x 6 = 3 The other 0.5 moles has 1.5Na which breaks into 3 atoms. So, total number of atoms = 0.5Na Now, degree of freedom for monoatomic gas = 3 So, degree of freedom for 0.5Na monoatomic particles = 1.5 x 0.5 = 0.75 Hence, total will be 3 + 0.75 = 3.75 which is the required answer. Thanks.

Thx very much . The answer is really very helpful

half of gas dissociated to become monoatomic, so its degree of freedom will be 
3 X 1/2 = 3/2=1.5
 and 50% are still as triatomic so there degree of freedom is;
1/2 X 6=3
 new degree of freedom of mixture is (1.5+3)/ 1/2+3/2=.4.5/2=3.75

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