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Byju's Answer

Standard XII

Physics

Equipartition Theorem

A sample of i...

Question

A sample of ideal gas undergoes Isothermal expansion in a reversible manner from volume $V_{1}$ to volume $V_{2}$ .
The initial pressure is $P_{1}$ and the final pressure is $P_{2}$ . The same sample is then allowed to undergo reversible expansion under adiabatic conditions from volume $V_{1}$ to $V_{2}$ . The initial pressure being same but final pressure is $P_{3}$ .What is the relationship between $P_{2}$ and $P_{3}$ ?

P_{3} = P_{2} {(\frac{V_{2}}{V_{1}})}^{γ}

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P_{3} = P_{2} {(\frac{V_{1}}{V_{2}})}^{γ - 2}

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P_{3} = P_{2} {(\frac{V_{1}}{V_{2}})}^{γ - 1}

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Solution

The correct option is C $P_{3} = P_{2} {(\frac{V_{1}}{V_{2}})}^{γ - 1}$
Since the first process is isothermal, we have
$P_{1} V_{1} = P_{2} V_{2}$
Now it is given that the same process is carried out adiabatically. Hence, we have,
$P_{1} V_{1}^{γ} = P_{3} V_{2}^{γ}$
$⟹ P_{3} = P_{1} (\frac{V_{1}^{γ}}{V_{2}^{γ}})$
But
$P_{1} = P_{2} (\frac{V_{2}}{V_{1}})$
Hence we get
$P_{3} = P_{2} {(\frac{V_{1}}{V_{2}})}^{γ - 1}$
Hence option C is the right answer.

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The correct option is C P3=P2(V1V2)γ−1Since the first process is isothermal, we haveP1V1=P2V2Now it is given that the same process is carried out adiabatically. Hence, we have,P1Vγ1=P3Vγ2⟹P3=P1(Vγ1Vγ2)But P1=P2(V2V1)Hence we getP3=P2(V1V2)γ−1Hence option C is the right answer.