1 mole of an ideal gas A(CV, m = 3R) and 2 mole of an ideal gas B are taken in a container and expanded reversible and adiabatically from 1 litre to 4 litre starting from initial temperature of 320 K. ΔE or ΔU for the process is :
Question
1 mole of an ideal gas A(CV, m = 3R) and 2 mole of an ideal gas B are taken in a container and expanded reversible and adiabatically from 1 litre to 4 litre starting from initial temperature of 320 K. ΔE or ΔU for the process is :
Solution
The change in internal energy (ΔU) for an adiabatic process is given by the formula:
ΔU = nCvΔT
where: n is the number of moles, Cv is the molar specific heat at constant volume, ΔT is the change in temperature.
Given: n1 = 1 mole (for gas A), Cv1 = 3R (for gas A), n2 = 2 moles (for gas B), Cv2 = Cv (for gas B, since it's not given we assume it to be Cv), Initial temperature (T1) = 320 K, Final volume = 4 litres, Initial volume = 1 litre.
We know that for an adiabatic process, T1V1^(γ-1) = T2V2^(γ-1), where γ = Cp/Cv.
For gas A, γ1 = Cp1/Cv1 = (Cv1 + R)/Cv1 = 4/3. For gas B, γ2 = Cp2/Cv2 = (Cv2 + R)/Cv2 = γ (since Cv2 = Cv).
Substituting the values in the adiabatic process formula, we get:
For gas A, 320 * (1)^(4/3-1) = T2 * (4)^(4/3-1) => T2 = 320 * (1/4)^(1/3) = 320/((2^(2/3)) * 2) = 320/(2^(8/3)) = 320/2^(2.67) = 320/6.3496042078727975 = 50.39 K
For gas B, 320 * (1)^(γ-1) = T2 * (4)^(γ-1) => T2 = 320 * (1/4)^(1/γ-1) = 320/(2^(2/γ)) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ) = 320/2^(2/γ
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