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Problem 3 A sample of 2.37 moles of an ideal diatomic gas experiences a temperature increase of 65.2 K at constant volume. (a) Find the increase in internal energy if only translational and rotational motions are possible. (b) Find the increase in internal energy if translational, rotational, and vibrational motions are possible. (c) How much of the energy calculated in (a) and (b) is translational kinetic energy?

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Answer:

a) the increase in internal energy is 3211.78 J

b) dU = 3854.14 J

c) dU[tex]_{T}[/tex] = 1927.06 J

Explanation:

Given the data in question;

Foe a diatomic gas, the degree of freedom are as follow;

lets consider the positional degree of freedom

transitional df = 3

rotational df = 2

vibrational ff = 1

now, the internal energy given by;

U = Nf Ă— 1/2NKT = NfĂ—1/2Ă—nRT

where Nf is the number of degree of freedom

N is Number of atoms or molecules

n = number of molecules

L is Boltzmann constant

R is universal gas constant

so change in internal energy , change in T is given by

dU =  Nf × 1/2 × nT dT

n = 2.37 moles

dT = 65.2 K

R = 8.314 J/mol.J

a)

Find the increase in internal energy if only translational and rotational motions are possible

since rotational and transitional motion are involved ;

Nf = 3(trasitional) + 2(rotational) = 5

so,

dU = 5 × 1/2  × nRdT

we substitute

dU = 5 × 0.5  × 2.37 × 8.314 × 65.2

dU = 3211.78 J

Therefore, the increase in internal energy is 3211.78 J

b)

Find the increase in internal energy if translational, rotational, and vibrational motions are possible.

Nf = 3 + 2 + 1 = 6

dU = 6 × 1/2  × nRdT

dU = 6 × 0.5  × 2.37 × 8.314× 65.2

dU = 3854.14 J

c)  

How much of the energy calculated in (a) and (b) is translational kinetic energy?

dU[tex]_{T}[/tex] = 3 × 0.5  × 2.37 × 8.314 × 65.2

dU[tex]_{T}[/tex] = 1927.06 J

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