Heat and Thermodynamics 5 Question 9

9. A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is $T V^{x}=$ constant, then $x$ is

(a) $\frac{2}{5}$

(b) $\frac{2}{3}$

(c) $\frac{5}{3}$

(d) $\frac{3}{5}$

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Answer:

Correct Answer: 9. (a)

Solution:

  1. Key Idea For an ideal gas undergoing an adiabatic process at room temperature,

$$ p V^{\gamma}=\text { constant or } T V^{\gamma-1}=\text { constant } $$

For a diatomic gas, degree of freedom, $f=5$

$$ \therefore \quad \gamma=1+2 / f=1+\frac{2}{5}=\frac{7}{5} $$

As for adiabatic process, $T V^{\gamma-1}= constant \cdots(i)$

and it is given that, here $T V^{x}= constant \cdots(ii)$

Comparing Eqs. (i) and (ii), we get

or

$$ \begin{gathered} \gamma-1=x \Rightarrow \frac{7}{5}-1=x \\ x=2 / 5 \end{gathered} $$



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