Q1: A vessel of volume 1000 m³ contains air which is saturated with water vapour. The total pressure and temperature are 100 kPa and 20°C respectively. Assuming that the vapour pressure of water at 20°C is 2.34 kPa, the amount of water vapour (in kg) in the vessel is approximately

ANS:A - 17

To find the amount of water vapor in the vessel, we can use the ideal gas law in conjunction with Dalton's law of partial pressures. Dalton's law states that in a mixture of non-reacting gases, the total pressure exerted by the mixture is equal to the sum of the partial pressures of individual gases. Given:

• Total pressure (𝑃totalPtotal​) = 100 kPa
• Temperature (𝑇T) = 20°C = 293.15 K (converted to Kelvin)
• Vapour pressure of water (𝑃waterPwater​) = 2.34 kPa

The pressure exerted by dry air (𝑃airPair​) can be calculated by subtracting the vapor pressure of water from the total pressure: 𝑃air=𝑃total−𝑃water=100 kPa−2.34 kPa=97.66 kPaPair​=Ptotal​−Pwater​=100kPa−2.34kPa=97.66kPa Now, we can use the ideal gas law to find the amount of dry air in the vessel: 𝑃𝑉=𝑛𝑅𝑇PV=nRT Where:

• 𝑃P is the pressure of the dry air
• 𝑉V is the volume of the vessel
• 𝑛n is the number of moles of dry air
• 𝑅R is the gas constant
• 𝑇T is the temperature

Rearranging the equation to solve for 𝑛n, we get: 𝑛=𝑃𝑉𝑅𝑇n=RTPV​ Now, we can calculate the number of moles of dry air: 𝑛=97.66 kPa×1000 m38.314 J/mol⋅K×293.15 Kn=8.314J/mol⋅K×293.15K97.66kPa×1000m3​ 𝑛≈3980.43 molesn≈3980.43moles Now, to find the mass of water vapor, we can use the fact that the vapor is saturated, meaning the amount of water vapor corresponds to the vapor pressure at that temperature. Using the ideal gas law for water vapor: 𝑃water𝑉=𝑛water𝑅𝑇Pwater​V=nwater​RT We rearrange it to solve for 𝑛waternwater​: 𝑛water=𝑃water𝑉𝑅𝑇nwater​=RTPwater​V​ Substituting the given values: 𝑛water=2.34 kPa×1000 m38.314 J/mol⋅K×293.15 Knwater​=8.314J/mol⋅K×293.15K2.34kPa×1000m3​ 𝑛water≈797.73 molesnwater​≈797.73moles Finally, to find the mass of water vapor (𝑚watermwater​), we use the molecular weight of water (𝑀𝑊waterMWwater​): 𝑚water=𝑛water×𝑀𝑊watermwater​=nwater​×MWwater​ 𝑚water=797.73 moles×0.018 kg/molmwater​=797.73moles×0.018kg/mol 𝑚water≈14.36 kgmwater​≈14.36kg Therefore, the amount of water vapor in the vessel is approximately 14.36 kg.