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) $\frac{3}{5}$
Show Answer
Answer:
Correct Answer: 9. (a)
Solution:
- 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
$$ \begin{gathered} \gamma-1=x \Rightarrow \frac{7}{5}-1=x \\ x=2 / 5 \end{gathered} $$
