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Standard XII

Physics

Degree of Freedom

A rigid diato...

Question

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:

\frac{3}{5}

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\frac{2}{5}

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\frac{2}{3}

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\frac{5}{3}

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Solution

The correct option is B $\frac{2}{5}$
The equation of adiabatic change is given by,

$T V^{γ - 1} = constant$

$For rigid diatomic gas, γ = \frac{7}{5}$

$\Rightarrow T V^{\frac{7}{5} - 1} = T V^{\frac{2}{5}} = constant$

$∴ x = \frac{2}{5}$

Hence, option $(B)$ is correct.

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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:

The correct option is B 25The equation of adiabatic change is given by, TVγ−1=constant For rigid diatomic gas, γ=75 ⇒TV75−1=TV25=constant ∴x=25 Hence, option (B) is correct.

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:

The correct option is B $\frac{2}{5}$
The equation of adiabatic change is given by,

$T V^{γ - 1} = constant$

$For rigid diatomic gas, γ = \frac{7}{5}$

$\Rightarrow T V^{\frac{7}{5} - 1} = T V^{\frac{2}{5}} = constant$

$∴ x = \frac{2}{5}$

Hence, option $(B)$ is correct.