A saturated parcel of air is cooled from a temperature where saturation vapor pressure is 3 kPa to a temperature where the saturation pressure is 1 kPa. The total pressure is 100 kPa throughout. If the air parcel contains 1 kg of dry air, how much water vapour (in grams) will condense out of this air during the process? (Answer to nearest integer) 6 gm 9 gm 13 gm 26 gm

To solve this problem, we need to understand that the amount of water vapor that will condense out of the air is equal to the difference in the saturation vapor pressures at the two temperatures.

1. The initial saturation vapor pressure is 3 kPa and the final saturation vapor pressure is 1 kPa. The difference between these two pressures is 3 kPa - 1 kPa = 2 kPa.

2. The total pressure is 100 kPa, but we are only interested in the part of the pressure that is due to the water vapor. This is the partial pressure of the water vapor, which is equal to the saturation vapor pressure. So, the partial pressure of the water vapor decreases by 2 kPa during the cooling process.

3. The partial pressure of the water vapor is related to the mass of the water vapor by the ideal gas law: P = ρRT, where P is the pressure, ρ is the density, R is the specific gas constant for water vapor, and T is the temperature. We can rearrange this equation to solve for the density: ρ = P/RT.

4. We can use this equation to find the change in the mass of the water vapor. The change in pressure is 2 kPa = 2000 Pa, the specific gas constant for water vapor is R = 461 J/(kg·K), and we'll assume that the temperature is approximately constant at T = 273 K (the freezing point of water). So, the change in the density of the water vapor is ρ = 2000 Pa / (461 J/(kg·K) * 273 K) = 0.008 kg/m^3.

5. Finally, we can find the change in the mass of the water vapor by multiplying the change in the density by the volume of the air parcel. The volume of 1 kg of dry air is approximately 0.84 m^3, so the change in the mass of the water vapor is 0.008 kg/m^3 * 0.84 m^3 = 0.00672 kg.

6. To convert this mass from kilograms to grams, we multiply by 1000: 0.00672 kg * 1000 = 6.72 g.

So, to the nearest integer, 7 grams of water vapor will condense out of the air during the cooling process. However, this option is not available in the choices given. The closest option available is 6 gm.

To solve this problem, we need to understand that the amount of water vapor that will condense out of the air is equal to the difference in the saturation vapor pressures at the two temperatures.

1. First, we find the initial amount of water vapor in the air. This is given by the initial saturation vapor pressure (3 kPa) divided by the total pressure (100 kPa), multiplied by the mass of the dry air (1 kg). This gives us 0.03 kg of water vapor.

2. Next, we find the final amount of water vapor in the air. This is given by the final saturation vapor pressure (1 kPa) divided by the total pressure (100 kPa), multiplied by the mass of the dry air (1 kg). This gives us 0.01 kg of water vapor.

3. The amount of water vapor that condenses out of the air is then the initial amount minus the final amount. This gives us 0.02 kg of water vapor.

4. Finally, we convert this mass from kilograms to grams by multiplying by 1000. This gives us 20 grams of water vapor.

So, the answer is 20 grams, which is not listed in the options. There might be a mistake in the question or the provided options.

To solve this problem, we need to understand that the amount of water vapor that will condense out of the air is equal to the difference in the saturation vapor pressures at the two temperatures.

1. First, we find the initial amount of water vapor in the air. This is given by the initial saturation vapor pressure (3 kPa) divided by the total pressure (100 kPa), multiplied by the mass of the dry air (1 kg). This gives us 0.03 kg of water vapor.

2. Next, we find the final amount of water vapor in the air. This is given by the final saturation vapor pressure (1 kPa) divided by the total pressure (100 kPa), multiplied by the mass of the dry air (1 kg). This gives us 0.01 kg of water vapor.

3. The amount of water vapor that condenses out of the air is then the initial amount minus the final amount. This gives us 0.03 kg - 0.01 kg = 0.02 kg.

4. Finally, we convert this mass from kilograms to grams by multiplying by 1000. This gives us 20 grams.

So, the amount of water vapor that will condense out of the air is 20 grams. However, this option is not available in the choices. It seems there might be a mistake in the question or the provided choices.