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  • For a gaseous expansion, the amount of work done is times less than the heat supp

For a gaseous expansion, the amount of work done is\mathrm{ n(>1)} times less than the heat supplied. The degree of freedom of a gas molecule is equal to,
 

Option: 1

\mathrm{n^n}

 


Option: 2

\mathrm{2 n-1}
 


Option: 3

\mathrm{n^{1 / 3}-1}
 


Option: 4

\mathrm{2(n-1)}


Answers (1)

best_answer

1st law of thermodynamics

\mathrm{ \Delta Q=W+\Delta U }

\mathrm{ \text { Since, W }=\Delta Q / n }

\mathrm{ \Delta Q=n(\Delta Q)+\Delta U }

\mathrm{ \Rightarrow \Delta Q[1-1 / n]=\Delta U }

\mathrm{ \Rightarrow m C_p \theta[1-1 / n]=m C_v \theta }

\mathrm{ \Rightarrow(1-1 / n)=\frac{C_v}{C_p}, \text { putting } \frac{C_p}{C_v}=\gamma \text { we obtain } 1-\frac{1}{n}=\frac{1}{\gamma} }

\mathrm{ Putting \: \gamma=1+\frac{2}{\mathrm{i}} where\: \mathrm{i}=\text{degree of freedom we obtain} }

\mathrm{ \quad \frac{n-1}{n}=\frac{i}{i+2} \Rightarrow n i+2 n-i-2=n i }

\mathrm{ \therefore I=2(n-1) }

\mathrm{ \therefore } option 4 is correct.














 

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Anam Khan

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