Nitrogen gas is filled in an insulated container. If α fraction of moles dissociates without exchange of any energy, then the fractional change in its temperature is

Question

Nitrogen gas is filled in an insulated container. If α fraction of moles dissociates without exchange of any energy, then the fractional change in its temperature is (A) (-α/5+α) (B) (α/3+α) (C) (-3 α/2+α) (D) (5 α/2+3 α)

Tardigrade Admin · Accepted Answer

Degree of freedom of diatomic nitrogen =5 Degree of freedom of monoatomic nitrogen =3 Let initial number of moles be n and α fraction dissociated. So fraction dissociated =n α fraction remaining =n-n α. nα break into two so new atoms formed is actually 2 nα. Initial energy is given by =n × (f/2) × R T=n × (5/2) × R T Final energy =(n-n α) (5/2) R T2+2 n α × (3/2) R T2 =(5/2) n R T2-(5/2) n α R T2+n α 3 R T2 =(5/2) n R T2+(n α R T2/2) =((5+2) n R T2/2) Change in energy is given on zero. (5 n R T/2)=((5+α) n R T2/2) (5 T/5+α)=T2 Δ T=T2-T or Δ T=(5 T/5+α)-T=(-α/5+α) T Fractional change in temperature =(Δ T/T) or -(α/5+α)

Q. Nitrogen gas is filled in an insulated container. If $α$ fraction of moles dissociates without exchange of any energy, then the fractional change in its temperature is

$\frac{- α}{5 + α}$

$\frac{α}{3 + α}$

$\frac{- 3 α}{2 + α}$

$\frac{5 α}{2 + 3 α}$

Q. Nitrogen gas is filled in an insulated container. If α fraction of moles dissociates without exchange of any energy, then the fractional change in its temperature is

5+α−α​

3+αα​

2+α−3α​

2+3α5α​

Q. Nitrogen gas is filled in an insulated container. If $α$ fraction of moles dissociates without exchange of any energy, then the fractional change in its temperature is

$\frac{- α}{5 + α}$

$\frac{α}{3 + α}$

$\frac{- 3 α}{2 + α}$

$\frac{5 α}{2 + 3 α}$