The pressure of a moist gas at 27°C is 4 atm. The volume of the container is doubled at the same temperature. The new pressure of the moist gas is .... × 10⁻¹ atm. (Nearest integer)
(Given: The vapour pressure of water at 27°C is 0.4 atm)
(Given: The vapour pressure of water at 27°C is 0.4 atm)
Answer: 22
Solution
Initially, the total pressure of moist gas = pressure of dry gas + vapor pressure of water
P_total(initial) = P_gas(initial) + P_water = 4 atm
P_gas(initial) = 4 - 0.4 = 3.6 atm
When volume is doubled at constant temperature, the pressure of the dry gas follows Boyle's law:
P₁V₁ = P₂V₂
3.6 × V = P_gas(new) × 2V
P_gas(new) = 3.6/2 = 1.8 atm
The vapor pressure of water remains constant at 0.4 atm (it depends only on temperature).
New total pressure = P_gas(new) + P_water = 1.8 + 0.4 = 2.2 atm = 22 × 10⁻¹ atm
Answer: 22
P_total(initial) = P_gas(initial) + P_water = 4 atm
P_gas(initial) = 4 - 0.4 = 3.6 atm
When volume is doubled at constant temperature, the pressure of the dry gas follows Boyle's law:
P₁V₁ = P₂V₂
3.6 × V = P_gas(new) × 2V
P_gas(new) = 3.6/2 = 1.8 atm
The vapor pressure of water remains constant at 0.4 atm (it depends only on temperature).
New total pressure = P_gas(new) + P_water = 1.8 + 0.4 = 2.2 atm = 22 × 10⁻¹ atm
Answer: 22
JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Integer answer