Question

unit 2 worksheet 2 - measuring pressure
problems 1 and 2. calculate the pressure of the gas in the flask connected to the manometer

1. proom = 730 mmhg

a. b. c.

1. proom = 733 mmhg

a. b. c.

1. what do we mean by atmospheric pressure? what causes this pressure?
2. how do we measure atmospheric pressure? is atmospheric pressure the same everywhere on the surface of the earth?

Explanation:

Step1: Determine pressure difference

The pressure of the gas \(P_g\) is found by considering the height - difference in the manometer. The liquid level is higher on the side open to the room, so the gas pressure is higher. The pressure difference \(\Delta P\) is the difference in the liquid - column heights. \(\Delta P=(127 - 64)\ mmHg\).
\(P_g=P_{room}+\Delta P\)
\(P_g = 730+(127 - 64)\)
\(P_g=730 + 63=793\ mmHg\)
- **b.**
- # Explanation:
## Step1: Determine pressure difference
The liquid level is higher on the side connected to the gas, so the gas pressure is lower. The pressure difference \(\Delta P=(130 - 26)\ mmHg\).
\(P_g=P_{room}-\Delta P\)
\(P_g = 730-(130 - 26)\)
\(P_g=730 - 104 = 626\ mmHg\)
- **c.**
- # Explanation:
## Step1: Note equal - level condition
Since the liquid levels in both arms of the manometer are the same, the gas pressure is equal to the room pressure.
\(P_g = P_{room}=730\ mmHg\)
2. **For problem 2 with \(P_{room}=733\ mmHg\):**
- **a.**
- # Explanation:
## Step1: Determine pressure difference
The liquid level is higher on the side open to the room, so the gas pressure is higher. The pressure difference \(\Delta P=(95 - 41)\ mmHg\).
\(P_g=P_{room}+\Delta P\)
\(P_g = 733+(95 - 41)\)
\(P_g=733 + 54=787\ mmHg\)
- **b.**
- # Explanation:
## Step1: Determine pressure difference
The liquid level is higher on the side connected to the gas, so the gas pressure is lower. The pressure difference \(\Delta P=(104 - 31)\ mmHg\).
\(P_g=P_{room}-\Delta P\)
\(P_g = 733-(104 - 31)\)
\(P_g=733 - 73 = 660\ mmHg\)
- **c.**
- # Explanation:
## Step1: Determine pressure difference
The liquid level is higher on the side connected to the gas, so the gas pressure is lower. The pressure difference \(\Delta P=(138 - 98)\ mmHg\).
\(P_g=P_{room}-\Delta P\)
\(P_g = 733-(138 - 98)\)
\(P_g=733 - 40 = 693\ mmHg\)
3. **Answer to question 3**:
- # Explanation:
## Step1: Define atmospheric pressure
Atmospheric pressure is the force per unit area exerted on a surface by the weight of the air above it in the atmosphere. It is caused by the gravitational pull on the air molecules in the atmosphere. The air molecules are in constant motion and collide with surfaces, exerting a force.
- # Answer:
Atmospheric pressure is the pressure exerted by the weight of the air in the Earth's atmosphere. It is caused by the gravitational attraction of the Earth on the air molecules.
4. **Answer to question 4**:
- # Explanation:
## Step1: Describe measurement method
Atmospheric pressure can be measured using a barometer. A mercury barometer consists of a long, evacuated tube inverted in a dish of mercury. The height of the mercury column in the tube is proportional to the atmospheric pressure. An aneroid barometer uses a sealed, flexible metal chamber (aneroid cell) that changes shape with changes in atmospheric pressure.
## Step2: Discuss uniformity
Atmospheric pressure is not the same everywhere on the Earth's surface. It varies with altitude (decreases as altitude increases), weather conditions (high - pressure systems have higher pressure than low - pressure systems), and temperature.
- # Answer:
Atmospheric pressure is measured using a barometer (mercury or aneroid). It is not the same everywhere on the Earth's surface; it varies with altitude, weather, and temperature.

Answer:

1. For problem 1 with \(P_{room}=730\ mmHg\):

• a.
• # Explanation:

Step1: Determine pressure difference

The pressure of the gas \(P_g\) is found by considering the height - difference in the manometer. The liquid level is higher on the side open to the room, so the gas pressure is higher. The pressure difference \(\Delta P\) is the difference in the liquid - column heights. \(\Delta P=(127 - 64)\ mmHg\).
\(P_g=P_{room}+\Delta P\)
\(P_g = 730+(127 - 64)\)
\(P_g=730 + 63=793\ mmHg\)

• b.
• # Explanation:

Step1: Determine pressure difference

The liquid level is higher on the side connected to the gas, so the gas pressure is lower. The pressure difference \(\Delta P=(130 - 26)\ mmHg\).
\(P_g=P_{room}-\Delta P\)
\(P_g = 730-(130 - 26)\)
\(P_g=730 - 104 = 626\ mmHg\)

• c.
• # Explanation:

Step1: Note equal - level condition

Since the liquid levels in both arms of the manometer are the same, the gas pressure is equal to the room pressure.
\(P_g = P_{room}=730\ mmHg\)

1. For problem 2 with \(P_{room}=733\ mmHg\):

• a.
• # Explanation:

Step1: Determine pressure difference

The liquid level is higher on the side open to the room, so the gas pressure is higher. The pressure difference \(\Delta P=(95 - 41)\ mmHg\).
\(P_g=P_{room}+\Delta P\)
\(P_g = 733+(95 - 41)\)
\(P_g=733 + 54=787\ mmHg\)

• b.
• # Explanation:

Step1: Determine pressure difference

The liquid level is higher on the side connected to the gas, so the gas pressure is lower. The pressure difference \(\Delta P=(104 - 31)\ mmHg\).
\(P_g=P_{room}-\Delta P\)
\(P_g = 733-(104 - 31)\)
\(P_g=733 - 73 = 660\ mmHg\)

• c.
• # Explanation:

Step1: Determine pressure difference

The liquid level is higher on the side connected to the gas, so the gas pressure is lower. The pressure difference \(\Delta P=(138 - 98)\ mmHg\).
\(P_g=P_{room}-\Delta P\)
\(P_g = 733-(138 - 98)\)
\(P_g=733 - 40 = 693\ mmHg\)

1. Answer to question 3:

• # Explanation:

Step1: Define atmospheric pressure

Atmospheric pressure is the force per unit area exerted on a surface by the weight of the air above it in the atmosphere. It is caused by the gravitational pull on the air molecules in the atmosphere. The air molecules are in constant motion and collide with surfaces, exerting a force.

• # Answer:

Atmospheric pressure is the pressure exerted by the weight of the air in the Earth's atmosphere. It is caused by the gravitational attraction of the Earth on the air molecules.

1. Answer to question 4:

• # Explanation:

Step1: Describe measurement method

Atmospheric pressure can be measured using a barometer. A mercury barometer consists of a long, evacuated tube inverted in a dish of mercury. The height of the mercury column in the tube is proportional to the atmospheric pressure. An aneroid barometer uses a sealed, flexible metal chamber (aneroid cell) that changes shape with changes in atmospheric pressure.

Step2: Discuss uniformity

Atmospheric pressure is not the same everywhere on the Earth's surface. It varies with altitude (decreases as altitude increases), weather conditions (high - pressure systems have higher pressure than low - pressure systems), and temperature.

• # Answer:

Atmospheric pressure is measured using a barometer (mercury or aneroid). It is not the same everywhere on the Earth's surface; it varies with altitude, weather, and temperature.