SATHEE: Heat and Thermodynamics 5 Question 9

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) $\frac{2}{3}$

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

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Solution:

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

$p V^{γ} = constant or T V^{γ - 1} = constant$

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

$∴ γ = 1 + 2 / f = 1 + \frac{2}{5} = \frac{7}{5}$

As for adiabatic process, $T V^{γ - 1} =$ constant

and it is given that, here $T V^{x} =$ constant

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

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

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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 TVx= constant, then x is

Solution:

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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