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Standard X

Chemistry

Gay Lussac's Law

A pressure co...

Question

A pressure cooker has the lid screwed on tight. A small opening of $10 m m^{2}$ is covered with a petcock that can be lifted to let steam escape. How much mass should the petcock have to draw boiling at $120^{\circ} C$ with an outside atmosphere at 101.3 kPa?
Take $P_{s a t} = 198.5 k P a (a t 120^{\circ} C)$

25 gm

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50 gm

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75 gm

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100 gm

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Solution

The correct option is D 100 gm
$F = m g = Δ P \times A$

$m = \frac{Δ P \times A}{g} = \frac{(198.5 - 101.3) \times 1000 \times 10^{-} 6 \times 10}{9.81}$

$= 99.08 ≃ 100 g m$

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A pressure cooker has the lid screwed on tight. A small opening of 10mm2 is covered with a petcock that can be lifted to let steam escape. How much mass should the petcock have to draw boiling at 120∘C with an outside atmosphere at 101.3 kPa? Take Psat=198.5kPa(at120∘C)

The correct option is D 100 gmF=mg=ΔP×A m=ΔP×Ag=(198.5−101.3)×1000×10−6×109.81 =99.08≃100gm

The correct option is D 100 gm
$F = m g = Δ P \times A$

$m = \frac{Δ P \times A}{g} = \frac{(198.5 - 101.3) \times 1000 \times 10^{-} 6 \times 10}{9.81}$

$= 99.08 ≃ 100 g m$