Question

An open-end manometer containing mercury is connected to a container of gas, as depicted in Sample Exercise 10.2. What is the pressure of the enclosed gas in torr in each of the following situations?
a) The mercury in the arm attached to the gas is 15.4 mm higher than in the one open to the atmosphere; atmospheric pressure is 0.985 atm.
b) The mercury in the arm attached to the gas is 12.3 mm lower than in the one open to the atmosphere; atmospheric pressure is 0.99 atm.

Answer 1

Answer:

For A  = 733.2  torr

For B = 740.1   torr

Explanation:

Given data:

For A

Height = 15.4 mm

Pressure of atmosphere = 0.985 atm ( 0.985 ×760 = 748.6  mmHg)

Pressure of gas = ?

For B

Height = 12.3 mm

Pressure of atmosphere = 0.99 atm ( 0.99×760 = 752.4 mmHg)

Pressure of gas = ?

Solution:

For A

P( gas) = P (atmosphere) - 15.4 mmHg

P( gas) = 748.6  mmHg - 15.4 mmHg

P( gas) = 733.2 mmHg or  733.2  torr

For B

P( gas) = P (atmosphere) - 12.3 mmHg

P( gas) = 752.4 mmHg - 12.3 mmHg

P( gas) =  740.1  mmHg

   or

740.1   torr

Answer 2

The pressure of the enclosed gas in torr in each of the following situations are as follows:

• A  = 733.2  torr
• B = 740.1   torr

What is Pressure?

This refers to the applied force to an object which is perpendicular to the area of the surface of the object.

Therefore, given

Height = 15.4 mm

Pressure of atmosphere = 0.985 atm ( 0.985 ×760 = 748.6  mmHg)

We need to find the pressure of gas

• P( gas) = P (atmosphere) - 15.4 mmHg
• P( gas) = 748.6  mmHg - 15.4 mmHg
• P( gas) = 733.2 mmHg or  733.2  torr

For question B

Height = 12.3 mm

Pressure of atmosphere = 0.99 atm ( 0.99×760 = 752.4 mmHg)

We need to find the Pressure of gas

• P( gas) = P (atmosphere) - 12.3 mmHg
• P( gas) = 752.4 mmHg - 12.3 mmHg
• P( gas) =  740.1  mmHg or 740.1   torr

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