Question

A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TV^x = constant, then x is
1. 2/5
2. 2/3
3. 5/3
4. 3/5

Answer 1

Correct Answer - Option 1 : 2/5

Concept:

Given

Relation between temperature and volume for the given adiabatic process is

TV^x = constant

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

pV^γ = 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 TV^x = 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 \(\frac{2}{5}.\)