Question 638: Thermodynamics & KTG

Question

When a diatomic gas (rigid) undergoes adiabatic change, its pressure $(\mathrm{P})$ and temperature $(\mathrm{T})$ are related as $P \propto T^c$. The value of $c$ is

Accepted Answer

For a
rigid diatomic gas
(meaning rotation is allowed but vibration is ignored), we proceed as per the NCERT method.
Step 1: Adiabatic Equation
For any adiabatic process:
$$ PV^\gamma = 	ext{constant} $$
where $\gamma$ is the adiabatic index.
Step 2: Ideal Gas Law
Also,
$$ PV = nRT $$
So,
$$ V = \frac{nRT}{P} $$
Step 3: Substitute $V$ in adiabatic equation
Substitute $V$ in the adiabatic equation:
$ P\left( \frac{nRT}{P} ight)^\gamma = 	ext{constant} $
$ P \cdot (nRT)^\gamma \cdot P^{-\gamma} = 	ext{constant} $
$ (nR)^\gamma \cdot P^{1-\gamma} \cdot T^\gamma = 	ext{constant} $
Since $n$ and $R$ are constants, we can absorb them into the overall constant:
$ P^{1-\gamma} \cdot T^\gamma = 	ext{constant} $
$ P^{1-\gamma} \propto T^{-\gamma} $
$ P \propto T^{\frac{\gamma}{\gamma - 1}} $
Step 4: Value of $\gamma$ for Diatomic Gas
For a diatomic gas (rotational but
not vibrational
):
Degrees of freedom $= 5$
$C_v = \frac{5}{2} R$
$C_p = C_v + R = \frac{7}{2}R$
$\gamma = \frac{C_p}{C_v} = \frac{7/2}{5/2} = \frac{7}{5} = 1.4$
Step 5: Value of $c$
$\displaystyle P \propto T^c\ \qquad	ext{where}\ c = \frac{\gamma}{\gamma - 1} $
$\gamma = \frac{7}{5}$
So,
$ c = \frac{\gamma}{\gamma - 1} = \frac{\frac{7}{5}}{\frac{7}{5} - 1} = \frac{\frac{7}{5}}{\frac{2}{5}} = \frac{7}{2} = 3.5 $
Correct answer: Option B (3.5)

Answer

2.5

Answer

3.5

Answer

1.5

Answer

5.2

Thermodynamics & KTG

When a diatomic gas (rigid) undergoes adiabatic change, its pressure (\mathrm{P}) and temperature (\mathrm{T}) are related as P \propto T^c. The value of c is

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