A manometer is connected to a gas containing bulb. the open arm reads $40.0$ cm whereas the arm connected to the bulb reads $15.0$ cm. If barometric pressure is $74.0$ cm Hg, then what is the pressure of a gas in the bar?

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Solution

excess pressure = $40 - 15 = 25 c m$
$∴$ total pressure of gas $= 74 + 25$
$= 99 c m$
$= 990 n m$
$760 n c m = 1$ bar
$∴ \frac{1}{760} \times 990 = 1.302$ Bar

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A manometer is connected to a gas containing bulb. the open arm reads 40.0 cm whereas the arm connected to the bulb reads 15.0 cm. If barometric pressure is 74.0 cm Hg, then what is the pressure of a gas in the bar?

excess pressure = 40−15=25cm∴ total pressure of gas =74+25 =99cm =990nm760ncm=1 bar∴1760×990=1.302 Bar

A manometer is connected to a gas containing bulb. the open arm reads $40.0$ cm whereas the arm connected to the bulb reads $15.0$ cm. If barometric pressure is $74.0$ cm Hg, then what is the pressure of a gas in the bar?

excess pressure = $40 - 15 = 25 c m$
$∴$ total pressure of gas $= 74 + 25$
$= 99 c m$
$= 990 n m$
$760 n c m = 1$ bar
$∴ \frac{1}{760} \times 990 = 1.302$ Bar