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The pressure (1 X 10⁵Nm_^-2) of the air filled in a vessel is decreased adiabatically so much as to increase its volume three times. The air pressure is ( $γ$ for air = 1.4, log₁₀ 3 = 0.4771, log₁₀ 2.148= 0.33206)

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1.14 X 10⁵ Nm^-2

2.14 X 10⁴ Nm^-2

2.19 x 10⁵ Nm-^-2

3.14 x 10⁴ Nm^-2

answer is A.

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Detailed Solution

Using $p_{1} V_{1}^{γ} = p_{2} V_{2}^{γ}, we get$

$p_{2} = p_{1} {(\frac{V_{1}}{V_{2}})}^{γ} = 1 \times 10^{5} {(\frac{1}{3})}^{14} = 2.14 \times 10^{4} {Nm}^{- 2}$

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