Three samples of the same gas A, B and C $(γ = 3 / 2)$ have equal volume initially. Now, the volume of each sample is doubled. For A, the process is adiabatic; for B, it is isobaric and for C, the process is isothermal. If the final pressures are equal for all the three samples, the ratio of their initial pressures is

2 : 1 : \sqrt{2}

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2 \sqrt{2} : 1 : 2

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\sqrt{2} : 1 : 2

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\sqrt{2} : 2 : 1

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Solution

The correct option is C $2 \sqrt{2} : 1 : 2$
Let the initial pressure of the three sample be $P_{A}$ .
$P_{B}$ and $P_{C}$
then $P_{A} (V)^{\frac{3}{2}} = (2 V)^{\frac{3}{2}} P$

$P_{B} = P$

$P_{C} (V) = P (2 V)$

$∴ P_{A} : P_{B} : P_{C} = (2)^{3 / 2}, 1 : 2 = 2 \sqrt{2} : 1 : 2$

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The correct option is C 2√2:1:2Let the initial pressure of the three sample be PA.PB and PC then PA(V)32=(2V)32P PB=PPC(V)=P(2V)∴PA:PB:PC=(2)3/2,1:2=2√2:1:2

The correct option is C $2 \sqrt{2} : 1 : 2$
Let the initial pressure of the three sample be $P_{A}$ .
$P_{B}$ and $P_{C}$
then $P_{A} (V)^{\frac{3}{2}} = (2 V)^{\frac{3}{2}} P$

$P_{B} = P$

$P_{C} (V) = P (2 V)$

$∴ P_{A} : P_{B} : P_{C} = (2)^{3 / 2}, 1 : 2 = 2 \sqrt{2} : 1 : 2$