Answer:
1. 0.9539 atm
2. 3112.732 mmHg
3. The third question is an incomplete one, as a diagram showing a gas tube with a measurement of the mercury-filled manometer should be included in the question. This will enable easy calculation of the pressure (in atmospheres) of the gas inside the container connected to an open-end, mercury-filled manometer. However, an example is explained to give a clear understanding on how it's been calculated.
Explanation:
1. On a rainy day, a barometer reads 725 mmHg. Convert this value to atmospheres.
The answer is:
1 atm=760 mmHg=101,325 Pa
so, 725 mmHg to atm =
725 mmHg ÷ 760 mmHg = 0.9539 atm
2. A closed container is filled with oxygen. The pressure in the container is 415 kPa. What is the pressure in millimeters of mercury?
The answer is :
1 atm=760 mmHg=101,325 Pa
First, 415 kPa should be converted to pa
1 kPa = 1000 pa
415 kPa = 415,000 pa
Afterwards, pa should be converted to atm
415,000 pa ÷ 101,325 Pa = 4.0957 atm
Finally, atm should be converted to mmHg
4.0957 × 760 = 3112.732 mmHg
3. What is the pressure (in atmospheres) of the gas inside the container connected to an open-end, mercury-filled manometer?
The answer is:
atmospheric pressure = density of mercury x acceleration due to gravity x height of column of mercury.
Let us use for instance, 20 cmHg as the pressure in the gas. To find pressure (in atmospheres) of the gas inside the container connected to an open-end, mercury-filled manometer, we have,
0.95atm + ( (20cmHg x 10mm) ÷ 1cm) ) x (1atm ÷ 760mmHg) = 1.21 atm
where, 0.95atm is the atmospheric pressure.
