1 ml of water vapour at STP is condensed to liquid calculate the decrease in volume.

To calculate the decrease in volume when 1 ml of water vapor at standard temperature and pressure (STP) is condensed into liquid water, we need to consider the properties of water in both its gaseous and liquid states. At STP, water vapor behaves as an ideal gas, and we can use the ideal gas law to help us with this calculation.

Understanding the Conditions

Standard temperature and pressure (STP) is defined as 0 degrees Celsius (273.15 K) and 1 atmosphere (101.3 kPa) of pressure. Under these conditions, 1 mole of an ideal gas occupies approximately 22.4 liters. Water vapor, being a gas at STP, will follow this behavior.

Volume of Water Vapor

When we have 1 ml of water vapor, we need to convert this volume into liters for consistency with the ideal gas law. Since 1 ml is equal to 0.001 liters, we can proceed with our calculations.

Calculating Moles of Water Vapor

Using the ideal gas law, we can find the number of moles of water vapor in 1 ml:

• Volume (V) = 0.001 L
• Pressure (P) = 1 atm
• Temperature (T) = 273.15 K

The ideal gas law is given by the equation:

PV = nRT

Where:

• P = pressure in atm
• V = volume in liters
• n = number of moles
• R = ideal gas constant (0.0821 L·atm/(K·mol))
• T = temperature in Kelvin

Rearranging the equation to solve for n (number of moles):

n = PV / RT

Plugging in the values:

n = (1 atm) * (0.001 L) / (0.0821 L·atm/(K·mol) * 273.15 K)

Calculating this gives:

n ≈ 0.0000446 moles

Volume of Liquid Water

Now, when this water vapor condenses into liquid water, we need to know the volume of the resulting liquid. The density of liquid water is approximately 1 g/cm³, which means that 1 gram of water occupies 1 ml. To find the mass of the condensed water, we can use the molar mass of water (approximately 18 g/mol):

mass = n × molar mass

Calculating the mass:

mass ≈ 0.0000446 moles × 18 g/mol ≈ 0.000803 grams

Since the density of water is 1 g/ml, the volume of the condensed water is:

volume = mass / density

volume ≈ 0.000803 grams / 1 g/ml ≈ 0.000803 ml

Determining the Decrease in Volume

Now we can find the decrease in volume by subtracting the volume of the liquid water from the volume of the water vapor:

Decrease in volume = Volume of vapor - Volume of liquid

Decrease in volume ≈ 1 ml - 0.000803 ml ≈ 0.999197 ml

Final Thoughts

In summary, when 1 ml of water vapor at STP is condensed into liquid water, the decrease in volume is approximately 0.9992 ml. This significant reduction highlights the difference in volume between gases and liquids, illustrating how gases occupy much more space than their liquid counterparts under the same conditions.