# A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TVx = c

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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
1. 2/5
2. 2/3
3. 5/3
4. 3/5

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Correct Answer - Option 1 : 2/5

Concept:

Given

Relation between temperature and volume for the given adiabatic process is

TVx = constant

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

pVγ = constant or TV(γ-1) = constant

Calculation:

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

$\therefore \gamma = 1 + \frac{2}{f}$

$\gamma = 1 + \frac{2}{5} = \frac{7}{5}$

As for adiabatic process, TV(γ-1) = constant      ----(i)

And it is given that, here TVx = constant      ----(ii)

Comparing equations (i) and (ii), we get

γ – 1 = x

$\frac{7}{5} - 1 = x$

$x = \frac{{7 - 5}}{5} = \frac{2}{5}\;$

Then, the value of x in the given relation is equal to $\frac{2}{5}.$