degree of fredo of 0.5 mole of triatomic gas . which is
undissociated = .05*6=3(6=degree of fredom of non linear triatomic gas )
degree of dissociated gas = 0.5/2*3=0.75(3=degree of f..of dissociated gas )
hence total degree of fredom =3+0.75=3.75
A linkage has 14 links and the number of loops is 5.
What is the value of degree of freedom of this linkage?
It is given that- a linkage has 14 links and number of loop is 5.
so, to calculate the degree of freedom , we have
where, F is the degree of freedom,
N is the links and
L is the no. of loops.
therefore, on substituting the value , we get
Hence, degree of freedom is 3.
it was the 3rd degree ves
The main idea has nothing to do with statics. While there are many possible explanations for degrees of freedom. Degree of fredom is the number of independent variables required to determine the answer is state.For typically,if you have a statistical model that you're trying to fit to some data,the degree.
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 + 0.73 = 3.75
which is the required answer.