During an adiabatic process, the pressure of a gas is proportiona… — Solution | JEE Physics

MCQMediumJEE 2024Isothermal & Adiabatic Processes

During an adiabatic process, the pressure of a gas is proportional to the cube of its absolute temperature. The ratio of $C_p/C_v$ is:

Answer

Correct answer:B

Standard Method

Given: During an adiabatic process, $P \propto T^3$ .

Find: The ratio $\gamma = \frac{C_p}{C_v}$ .

For an adiabatic process, the relation between pressure and temperature is

PT^{-\frac{\gamma}{\gamma-1}} = \text{constant}

Given $P \propto T^3$ , we write

P = kT^3

So the temperature exponent must satisfy

3 - \frac{\gamma}{\gamma-1} = 0

Therefore,

3 = \frac{\gamma}{\gamma-1}

3(\gamma-1) = \gamma

3\gamma - 3 = \gamma

2\gamma = 3

\gamma = \frac{3}{2}

Hence,

\frac{C_p}{C_v} = \frac{3}{2}

Therefore, the correct option is B.

Using adiabatic relation with ideal gas law

Given: $P \propto T^3$ , so

PT^{-3} = \text{constant}

Find: $\gamma = \frac{C_p}{C_v}$ .

From the adiabatic relation,

PV^\gamma = \text{constant}

Using the ideal gas law,

V = \frac{nRT}{P}

Substitute into the adiabatic equation:

P\left(\frac{nRT}{P}\right)^\gamma = \text{constant}

Ignoring fixed constants,

P^{1-\gamma}T^\gamma = \text{constant}

Since $P \propto T^3$ , compare powers of $T$ to obtain

\frac{\gamma}{1-\gamma} = -3

So,

\gamma = -3 + 3\gamma

3 = 2\gamma

\gamma = \frac{3}{2}

Therefore, the ratio $\frac{C_p}{C_v}$ is $\frac{3}{2}$ , so the correct option is B.

Using the adiabatic relation incorrectly as $PV=\text{constant}$ . That applies only to isothermal conditions for an ideal gas, not to an adiabatic process. Use $PV^\gamma=\text{constant}$ or the equivalent pressure-temperature form instead.
Equating the exponent of $T$ with the wrong sign. In the relation $PT^{-\frac{\gamma}{\gamma-1}}=\text{constant}$ , careful handling of the negative exponent is essential. Write the exponent equation explicitly before solving.
Confusing $\gamma$ with $C_v/C_p$ instead of $C_p/C_v$ . The required ratio is $\gamma=\frac{C_p}{C_v}$ , so after solving for $\gamma$ do not invert the answer.

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