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  • If n is the number of degrees of freedom of the molecules of an ideal gas. Then the ratio of <img alt="\frac{c_p}{c_v}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7Bc_p%7D%7B

If n is the number of degrees of freedom of the molecules of an ideal gas. Then the ratio of \frac{c_p}{c_v} is-

Option: 1

1+\frac{1}{n}


Option: 2

1+\frac{2}{n}


Option: 3

2+\frac{2}{n}


Option: 4

1+\frac{3}{n}


Answers (1)

best_answer

Average kinetic energy of gas molecules = n.\frac{1}{2}KT [K=Boltzmann constant]

As the molecules have no potential energy, the total internal energy of 1 mole of gas is-

\begin{aligned} &E=N_0 n \cdot \frac{1}{2} k T=\frac{n}{2} N_0 k T=\frac{n}{2} R T\left[N_0 k=R\right]\\ &\begin{aligned} & \therefore C_V=\frac{d E}{d t}=\frac{n}{2} R \\ & \therefore C_p=C_V+R=\left(\frac{n}{2}+1\right) R \\ & \text { Then, } \frac{C_p}{C_V}=\frac{\frac{n}{2}+1}{n / 2}=1+\frac{n}{n} \end{aligned} \end{aligned}

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Divya Prakash Singh

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