The vapor pressure of water at 20c is 17.5 mmhg if the pressure of a gas collected over water was measured to be 453.0 mmHg.
What is the pressure pf the pure gas?
The correct answer and explanation is :
To determine the pressure of the pure gas, we need to subtract the vapor pressure of water from the total pressure of the gas collected over water.
Given:
- Total pressure of the gas collected = 453.0 mmHg
- Vapor pressure of water at 20°C = 17.5 mmHg
The formula to calculate the pressure of the pure gas is:
[
P_{\text{pure gas}} = P_{\text{total}} – P_{\text{water vapor}}
]
[
P_{\text{pure gas}} = 453.0 \text{ mmHg} – 17.5 \text{ mmHg}
]
[
P_{\text{pure gas}} = 435.5 \text{ mmHg}
]
Explanation:
When a gas is collected over water, it gets mixed with water vapor because water evaporates at all temperatures, exerting its own vapor pressure. The total pressure measured includes both the gas and the water vapor.
To find the actual pressure of the collected gas, we must subtract the contribution of water vapor. In this case, at 20°C, the vapor pressure of water is 17.5 mmHg, meaning that part of the measured 453.0 mmHg comes from evaporated water.
By removing the vapor pressure of water from the total pressure, we obtain the true pressure of the dry gas, which is 435.5 mmHg. This is an important step in gas collection experiments since the presence of water vapor can affect calculations related to gas laws, such as the ideal gas law or Dalton’s Law of Partial Pressures.
This calculation is useful in laboratory settings, especially in determining the properties of gases, ensuring accurate stoichiometric calculations, and understanding gas behavior under different conditions.