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(D) According to the law of equipartition of energy, each degree of freedom contributes $\frac{1}{2} k_B T$ to the average energy of a molecule.For a

Vedclass · Accepted Answer

$925$

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(D) According to the law of equipartition of energy, each degree of freedom contributes $\frac{1}{2} k_B T$ to the average energy of a molecule.For a solid, each atom acts as a 3D harmonic oscillator, having $3$ degrees of freedom for kinetic energy and $3$ degrees of freedom for potential energy, totaling $6$ degrees of freedom.The average energy per atom is $U = 6 	imes \frac{1}{2} k_B T = 3 k_B T$.For $1 \, 	ext{mole}$ of substance, the internal energy is $U_m = 3 R T$.The molar specific heat capacity is $C_v = \frac{dU_m}{dT} = 3 R$.Given $R = 8.314 \, J \, mol^{-1} \, K^{-1}$ and atomic weight $M = 27 	imes 10^{-3} \, kg/mol$, the specific heat capacity $c$ is given by $c = \frac{C_v}{M} = \frac{3 R}{M}$.Substituting the values: $c = \frac{3 	imes 8.314}{27 	imes 10^{-3}} \approx \frac{24.942}{0.027} \approx 923.77 \, J \, kg^{-1} \, K^{-1}$.Rounding to the nearest provided option, we get $925 \, J \, kg^{-1} \, K^{-1}$.

Using the law of equipartition of energy, the specific heat (in $J\, kg^{-1}\, K^{-1}$) of aluminium at room temperature can be estimated to be (atomic weight of aluminium $= 27$).

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