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

MCQMediumJEE 2024Isothermal & Adiabatic Processes

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

A
$\frac{5}{3}$
B
$\frac{3}{2}$
C
$\frac{7}{5}$
D
$\frac{9}{7}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

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

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

For an adiabatic process, the pressure-temperature relation is

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

Since $P \propto T^3$ , write

P = kT^3

for some constant $k$ . Comparing powers of $T$ in the adiabatic relation:

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

Now solve for $\gamma$ :

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

3(\gamma - 1) = \gamma

3\gamma - 3 = \gamma

2\gamma = 3

\gamma = \frac{3}{2}

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

Using adiabatic relation with ideal gas law

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

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

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

Start with the adiabatic relation:

PV^\gamma = \text{constant}

Using the ideal gas law,

V = \frac{nRT}{P}

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

Ignoring fixed constants $nR$ , this becomes

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

Using $P \propto T^3$ , equate exponents to obtain

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

Solving,

\gamma = -3 + 3\gamma

3 = 2\gamma

\gamma = \frac{3}{2}

Hence, the correct answer is $\frac{3}{2}$ , that is, option B.

Common mistakes

Using the relation $PV^\gamma = \text{constant}$ directly without converting it into a pressure-temperature relation. This is wrong because the question gives a relation between $P$ and $T$ , not $P$ and $V$ . First express the adiabatic condition in terms of $P$ and $T$ .
Equating the proportionality $P \propto T^3$ with $P = T^3$ and then forgetting that only exponents matter. The numerical constant does not affect the exponent comparison. Focus on matching the power of $T$ correctly.
Making an algebra mistake while solving $3 = \frac{\gamma}{\gamma - 1}$ . This leads to a wrong value of $\gamma$ . Cross-multiply carefully and simplify step by step to get $2\gamma = 3$ .

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