A soft plastic bottle, filled with water of density $1 gm/cc$ , carries an inverted glass test-tube with some air (ideal gas) trapped as shown in the figure. The test-tube has a mass of $5 gm$ , and it is made of a thick glass of density $2.5 gm/cc$ . Initially, the bottle is sealed at atmospheric pressure $p_{0} = 10^{5} Pa$ so that the volume of the trapped air is $v_{0} = 3.3 cc$ . When the bottle is squeezed from outside at constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure $p_{0} + Δ p$ without changing its orientation. At this pressure, the volume of the trapped air is $v_{0} - Δ v$ .

Let $Δ v = X cc$ and $Δ p = Y \times 10^{3} Pa$ .

The value of $Y$ is

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Solution

Isothermal process for air,

$p_{1} v_{1} = p_{2} v_{2}$

$\Rightarrow 10^{5} (3.3) = p_{2} (3)$

$\Rightarrow p_{2} = 1.1 \times 10^{5} Pa$

So,

$Δ p = 0.1 \times 10^{5} Pa = 10 \times 10^{3} Pa$

$∴ Y = 10$

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Isothermal process for air, p1v1=p2v2 ⇒105(3.3)=p2(3) ⇒p2=1.1×105 Pa So, Δp=0.1×105 Pa=10×103 Pa ∴Y=10

Isothermal process for air,

$p_{1} v_{1} = p_{2} v_{2}$

$\Rightarrow 10^{5} (3.3) = p_{2} (3)$

$\Rightarrow p_{2} = 1.1 \times 10^{5} Pa$

So,

$Δ p = 0.1 \times 10^{5} Pa = 10 \times 10^{3} Pa$

$∴ Y = 10$