contestada

As illustrated, the below manometer consists of a gas vessel and an open-ended U-tube containing a nonvolatile liquid with a density of 0.993 g/mL. The difference in heights of the liquid in the two sides of the manometer is 32.3mm when the atmospheric pressure is 765 mm Hg. Given that the density of mercury is 13.6 g/mL, the pressure of the enclosed gas is ________ atm. Group of answer choices

Respuesta :

jufsanabriasa

Answer:

1.01atm is the pressure of the gas

Explanation:

The difference in heights in the two sides is because of the difference in  pressure of the enclosed gas and the atmospheric pressure. This difference is in mm of the nonvolatile liquid. The difference in mm Hg is:

32.3mm * (0.993g/mL / 13.6g/mL) = 2.36mmHg

As atmospheric pressure is 765mm Hg and assuming the gas has more pressure than the atmospheric pressure (There is no illustration), the pressure of the gas is:

765mm Hg + 2.36mm Hg = 767.36 mmHg

In atm:

767.36 mmHg * (1atm / 760 mmHg) =

1.01atm is the pressure of the gas

Cricetus

The pressure of the enclosed gas will be "0.993 atm".

Given:

  • Density of mercury = 13.6 g/mL
  • Density of nonvolatile liquid = 0.993 g/mL
  • Difference in height = 32.3mm
  • Atmospheric pressure = 765 mm Hg

Now,

The density of Hg will be:

= [tex]\frac{13.6}{0.791}[/tex]

= [tex]17.2 \ mm[/tex]

then,

= [tex]32.3\times (\frac{1 \ mm \ Hg}{17.2} )[/tex]

= [tex]18.73 \ mm \ Hg[/tex]

hence,

The pressure will be:

= [tex]765-18.73[/tex]

= [tex]746.27 \ mm \ Hg[/tex]

= [tex]0.993 \ atm[/tex]

Thus the above answer is right.

Learn more about gas vessel here:

https://brainly.com/question/7205294