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The vapour pressure of water at 300 K in a closed container is 0.4 atm. If the volume of the container is doubled, its vapour pressure at 300 K will be
- a)0.8 atm
- b)0.2 atm
- c)0.4 atm
- d)0.6 atm
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The vapour pressure of water at 300 K in a closed container is 0.4 atm...
The vapour pressure of water will remain same as the temperature is unchanged. So, the answer is c) 0.4 atm
Most Upvoted Answer
The vapour pressure of water at 300 K in a closed container is 0.4 atm...
Given:
Vapour pressure of water at 300 K = 0.4 atm
Volume of container = V
To find:
Vapour pressure of water at 300 K when the volume of container is doubled
Solution:
According to the Raoult's law, the vapour pressure of a liquid in a closed container is directly proportional to the mole fraction of the liquid present in the solution.
P_A = X_A.P°_A
Where,
P_A = Vapour pressure of water
X_A = Mole fraction of water
P°_A = Vapour pressure of pure water
Since the container is closed, the total pressure inside the container is the sum of the partial pressure of water vapour and the pressure due to the remaining gases.
Total pressure = P_A + P_g
where,
P_g = Pressure due to the remaining gases
As the volume of the container is doubled, the pressure due to the remaining gases remains constant. Therefore, the total pressure inside the container also remains constant.
P_A + P_g = constant
Therefore, the vapour pressure of water remains constant at 0.4 atm when the volume of the container is doubled.
Hence, the correct option is (c) 0.4 atm.
Vapour pressure of water at 300 K = 0.4 atm
Volume of container = V
To find:
Vapour pressure of water at 300 K when the volume of container is doubled
Solution:
According to the Raoult's law, the vapour pressure of a liquid in a closed container is directly proportional to the mole fraction of the liquid present in the solution.
P_A = X_A.P°_A
Where,
P_A = Vapour pressure of water
X_A = Mole fraction of water
P°_A = Vapour pressure of pure water
Since the container is closed, the total pressure inside the container is the sum of the partial pressure of water vapour and the pressure due to the remaining gases.
Total pressure = P_A + P_g
where,
P_g = Pressure due to the remaining gases
As the volume of the container is doubled, the pressure due to the remaining gases remains constant. Therefore, the total pressure inside the container also remains constant.
P_A + P_g = constant
Therefore, the vapour pressure of water remains constant at 0.4 atm when the volume of the container is doubled.
Hence, the correct option is (c) 0.4 atm.
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Community Answer
The vapour pressure of water at 300 K in a closed container is 0.4 atm...
The pressure exerted by vapour over the surface of a liquid at a given temperature is called vapour pressure ...... it (v .p) increases exponentially with increase in temperature....i.e, vapour pressure(v.p) is directly proportional to temperature...(v .p)1/(v .P)2=T1/T2.0.4/(v .p)2=300k/300k.0.4/(v .p)2= 1/1.(v .p)= 0.4..
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The vapour pressure of water at 300 K in a closed container is 0.4 atm. If the volume of thecontainer is doubled, its vapour pressure at 300 K will bea)0.8 atmb)0.2 atmc)0.4 atmd)0.6 atmCorrect answer is option 'C'. Can you explain this answer?
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