Question

a mixture of ideal gases has 1 mol of Ne, 2moles of N2 and 3moles of ozone at absolute temperature T. the internal energy of the mixture is​

Answer 1

Answer:

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Answer 2

The internal energy of the mixture is 31/2 RT.

Given-

• Moles of Neon = 1 mol
• Moles of nitrogen = 2 moles
• Moles of ozone = 3 moles

We know that internal energy associated with a gas having degree of freedom f and moles μ at absolute temperature T is given by-

U = f/2 μRT

where R is universal gas constant and T is temperature.

For 1 mole of Neon, f = 3 and μ = 1 mole

U₁ = 3/2 RT

For 2 moles of N₂, f = 5 and μ = 2 moles

U₂ = 5 RT

For 3 moles of O₃, f = 6 and μ = 3 moles

U₃ = 18/2 RT

So, total internal energy is

U = U₁ + U₂ + U₃

U = 3/2 RT + 5 RT + 18/2 RT

U = 31/2 RT