The initial pressure and volume of an ideal gas are P1 and V1. Th… - Solution | JEE Physics

MCQEasyJEE 2023Isothermal & Adiabatic Processes

The initial pressure and volume of an ideal gas are $P_1$ and $V_1$ . The final pressure of the gas when the gas is suddenly compressed to volume $\frac{V_1}{4}$ will be:

A
$P_1 \left( \frac{V_1}{4} \right)^\gamma$
B
$4 P_1$
C
$P_0$
D
$P_1 \left( \frac{V_1}{4} \right)^{\gamma-1}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Initial pressure and volume are $P_1$ and $V_1$ . The gas is suddenly compressed to volume $\frac{V_1}{4}$ .

Find: Final pressure of the gas.

Since the compression is sudden, the process is treated as adiabatic. Therefore,

PV^\gamma = \text{constant}

Using the relation between initial and final states as shown in the solution,

P_0 V_0^\gamma = P_2 \left(\frac{V_0}{4}\right)^\gamma

Rearranging,

P_2 = P_0 \cdot \frac{V_0^\gamma}{\left(\frac{V_0}{4}\right)^\gamma}

Now simplify:

P_2 = P_0 \cdot \frac{V_0^\gamma}{\frac{V_0^\gamma}{4^\gamma}} = P_0 \cdot 4^\gamma

Thus,

P_2 = P_0 (4)^\gamma

Therefore, the final pressure is $P_0 4^\gamma$ . The solution states the correct option is C, although this does not match the listed options verbatim.

Common mistakes

Treating the sudden compression as an isothermal process is incorrect because there is no time for heat exchange. Use the adiabatic relation $PV^\gamma = \text{constant}$ instead.
Substituting the final volume $\frac{V_1}{4}$ without raising the entire bracket to the power $\gamma$ gives a wrong pressure dependence. Apply the exponent to the complete volume term.
Confusing the symbols $P_1, V_1$ from the question with $P_0, V_0$ used in the solution can cause mapping errors. Keep track of initial and final states consistently before comparing with options.

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Answer

Step-by-step solution

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