Question

Two systems have the following equations of state, where RR is the gas constant per mole. The mole numbers are N_1N1​ = 3 and N_2N2​ = 5. The initial temperatures are T_1T1​ = 175 K and T_2T2​ = 400 K. What is the temperature once equilibrium is reached (in Kelvin)?

\frac{1}{T_1} = \frac{3}{2}R\frac{N_1}{U_1}T1​1​=23​RU1​N1​​

\frac{1}{T_2} = \frac{5}{2}R\frac{N_2}{U_2}T2​1​=25​RU2​N2​​

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Answer 1

Answer:

Plot the rotational partition function of N_22 as a function of temperature from 10 to 300 K. At what temperature (in K) does the approximation of q_r=\frac{T}{\Theta_r}qr=ΘrT (Eq. 5.49) result in less than 1% error?

Answer 2

Given info : no of moles of gases ; n₁ = 3 , n₂ = 5 initial temperature of gases ; T₁ = 175K , T₂ = 400K. Here 1/T₁ = (3/2)(Rn₁/U₁) means, U₁ = 3/2 n₁RT₁ similar, 1/T₂ = (5/2)(Rn₂/U₂) means, U₂ = 5/2n₂RT₂

We have to find the temperature once equilibrium is reached.

Solution : first find equilibrium molar heat capacity, Ceq = (n₁C₁ + n₂C₂)/(n₁ + n₂)

Here, C₁ = 3/2R , C₂ = 5/2R

so, Ceq = (3 × 3R/2 + 5 × 5R/2)/(3 + 5)

= (9R + 25R)/16

= 17R/8

Now from energy conservation theorem,

Initial energy of 1st and 2nd gases = energy of gases in equilibrium stage

⇒U₁ + U₂ = U

⇒3/2 n₁RT₁ + 5/2n₂RT₂ = nCeqT

⇒3/2 n₁RT₁ + 5/2n₂RT₂ = (3 + 5) 17R/8 T

⇒3/2 × 3 × 25/3 × 175 + 5/2 × 5 × 25/3 × 400 = 17 × 25/3 × T

⇒T = 340.44K

Therefore temperature once equilibrium is reached, is 340.44K