Adiabatic Compression Pressure Ratio Calculation

❓ Question

A monatomic gas having:

$$\gamma=\frac{5}{3}$$

is stored in a thermally insulated container and the gas is suddenly compressed to:

$$\frac{V}{16}$$

of its initial volume.

The ratio of pressure and volume represented by:

$$\frac{P_f}{P_i}$$

and

$$\frac{V_f}{V_i}$$

is to be determined.

Find the ratio of specific heats at constant pressure and constant volume.

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Adiabatic process
👉 Poisson’s equation
👉 Specific heat ratio.

Main idea:

For thermally insulated system:

$$PV^\gamma=\text{constant}$$

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🔷 Step 1 — Identify the Process 💯

Since container is thermally insulated:

$$Q=0$$

Hence process is:

$$\text{Adiabatic}$$

For adiabatic process:

$$PV^\gamma=\text{constant}$$

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🔷 Step 2 — Use Adiabatic Relation

$$P_iV_i^\gamma=P_fV_f^\gamma$$

Rearranging:

$$\frac{P_f}{P_i} = \left(\frac{V_i}{V_f}\right)^\gamma$$

Given:

$$V_f=\frac{V_i}{16}$$

So:

$$\frac{V_i}{V_f}=16$$

Thus:

$$\frac{P_f}{P_i}=16^\gamma$$

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🔷 Step 3 — Substitute $\gamma=\frac53$

$$\frac{P_f}{P_i} = 16^{5/3}$$

Now:

$$16=2^4$$

Hence:

$$16^{5/3} = (2^4)^{5/3} = 2^{20/3}$$

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🔷 Step 4 — Ratio Calculation

Required ratio:

$$\frac{\ln(P_f/P_i)}{\ln(V_i/V_f)}$$

Using:

$$\frac{P_f}{P_i}=\left(\frac{V_i}{V_f}\right)^\gamma$$

Taking logarithm:

$$\ln\left(\frac{P_f}{P_i}\right) = \gamma \ln\left(\frac{V_i}{V_f}\right)$$

Therefore:

$$\boxed{\gamma=\frac53}$$

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🔷 Step 5 — Evaluate Final Value

$$\gamma=\frac53$$
$$\gamma=\frac{5}{3}\approx1.67$$

Closest option:

$$\boxed{1.6}$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ Isothermal formula use kar lena

❌ $PV=\text{constant}$assume kar lena

❌ $V_f/V_i$ aur $V_i/V_f$ interchange kar dena

Remember:

$$\boxed{ PV^\gamma=\text{constant} }$$

for adiabatic process.

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✅ Final Answer

$$\boxed{1.6}$$

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