How far can I move (in cm) the piston of cross-section A1, in the device shown in figure, so that 40% of the air enclosed remains. (The atmospheric pressure p0 = 10 N/cm2, density of liquid = 5 g/cm3, ℓ = 20cm, A1 = 4cm2, A2 = 1 cm2. The temperature is constant. The piston is massless and the initial height of gas in device is 110 cm.

Engineering

Physics

Pressure Derivation Ideal Gas Equation and Gas Laws

Liquid Pressure

BernoulIis Equation and Equation of Continuity

Question

How far can I move (in cm) the piston of cross-section A₁, in the device shown in figure, so that 40% of the air enclosed remains. (The atmospheric pressure p₀ = 10 N/cm², density of liquid = 5 g/cm³, ℓ = 20cm, A₁= 4cm², A₂ = 1 cm². The temperature is constant. The piston is massless and the initial height of gas in device is 110 cm.

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$\frac{P_{0} V_{0}}{n_{0}} = \frac{P_{1} V_{1}}{n_{1}}$
$\frac{10^{5} \times A_{1} h_{0}}{n_{0}} = \frac{10^{5} + 5 \times 10^{3} \times 10 \times 0 .2 \times V_{1}}{0 .4 n_{0}}$
$\frac{10^{5} \times 4 \times 10^{- 4} \times h_{0}}{n_{0}} = \frac{(10^{5} + 0 .2 \times 10 \times 5 \times 10^{3}) V_{1}}{0 .4 n_{0}}$
   V₁ = 4 × 10^–4 [h₀ – h] + 0.2 × 1 × 10^–4
   h₀= 1.1 × 10⁵ × [4 × 10–4 (h₀– h) + 2 × 10^–5]
   = 1.1 [40h₀ – 40h + 2]
   16h₀ = 44h₀ – 44h + 2.2
   44h = 28h₀+ 2.2
   44h = 28 × 1.1 + 2.2 = 33
   h = 0.75 m