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

Answer 1

formula of internal energy at temperature T and no of mole of gas n is $nC_vT$ .
where Cv is molar heat capacity at constant volume .
N₂ is diatomic molecule so, Cv = 5R/2

Let initial temperature is T₁ and number of mole of N₂ is n
Then, internal energy { E } = n × 5R/2 × T₁
⇒T₁ = 2E/5nR -------(1)

after dissociation of N₂ gas ;
N₂ ----------> 2N
at t = 0
number of mole of N₂ = n
number of moles of N = 0

After dissociation by α fractional ,
number of moles of N₂ gas = n(1 - α)
number of moles of N = 2nα

Now, Let temperature be T₂ after dissociation ,
Then, internal energy { E } = energy of N₂ + energy of N
E = n(1 - α) × 5R/2 × T₂ + 2nα × 3R/2 × T₂ [ N is monoatomic so, Cv = 3R/2]
E = T₂nR/2[ 5(1 - α) + 6α ] = nRT₂[5 + α]/2
T₂ = 2E/nR[5 + α] ------(2)

Now, fractional change in temperature is $\bold{\frac{T_2-T_1}{T_1}}$
Fraction change = {2E/nR[5 + α] - 2E/5nR }/{2E/5nR}
= {1/(5+α) - 1/5}/{1/5}
= -α/(5 + α)

Hence, fractional change in temperature is -α/(5 + α)

Answer 2

Let initial temperature is T₁ and number of mole of N₂ is n

Then, internal energy { E } = n × 5R/2 × T₁

⇒T₁ = 2E/5nR -------(1)

after dissociation of N₂ gas ;

N₂ ----------> 2N

at t = 0

number of mole of N₂ = n

number of moles of N = 0

After dissociation by α fractional ,

number of moles of N₂ gas = n(1 - α)

number of moles of N = 2nα

Now, Let temperature be T₂ after dissociation ,

Then, internal energy { E } = energy of N₂ + energy of N

E = n(1 - α) × 5R/2 × T₂ + 2nα × 3R/2 × T₂ [ N is monoatomic so, Cv = 3R/2]

E = T₂nR/2[ 5(1 - α) + 6α ] = nRT₂[5 + α]/2

T₂ = 2E/nR[5 + α] ------(2)

Now, fractional change in temperature is

Fraction change = {2E/nR[5 + α] - 2E/5nR }/{2E/5nR}

= {1/(5+α) - 1/5}/{1/5}

= -α/(5 + α)

Hence, fractional change in temperature is -α/(5 + α)