Question 719: Thermodynamics & KTG

Question

For ideal non-rigid diatomic gas, the value of $\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}}$ is nearly $\left(\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{9}{7}\right)$

Accepted Answer

Given:
For an ideal non-rigid diatomic gas, $\gamma = \dfrac{C_P}{C_V} = \dfrac{9}{7}$
We are to find the value of $\dfrac{R}{C_V}$.
Step 1: Formula for $\gamma$
$ \gamma = \frac{C_P}{C_V} $
Step 2: Relation between $C_P$, $C_V$, and $R$
$ C_P - C_V = R \implies C_P = C_V + R $
Step 3: Plug this value into the formula for $\gamma$
$ \gamma = \frac{C_V + R}{C_V} $
$ \gamma = 1 + \frac{R}{C_V} $
Step 4: Rearranging to find $\frac{R}{C_V}$
$ \gamma = 1 + \frac{R}{C_V} $
$ \gamma - 1 = \frac{R}{C_V} $
Step 5: Substitute the value of $\gamma$
$ \gamma = \frac{9}{7} $
$ \gamma - 1 = \frac{9}{7} - 1 = \frac{9 - 7}{7} = \frac{2}{7} $
Therefore,
$ \boxed{\frac{R}{C_V} = \frac{2}{7} \approx 0.28} $
So, the correct option is
Option C: $0.28$

Answer

0.4

Answer

0.66

Answer

0.28

Answer

1.28

Thermodynamics & KTG

For ideal non-rigid diatomic gas, the value of \frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}} is nearly \left(\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{9}{7}\right)

Choose the correct answer: