Answer
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Hint: The ideal gas equation states that $PV = nRT$. The temperature is constant so the process is isothermal. The right hand side of the equation becomes constant as n, T and R are constant so,
\[PV = const\]
So, the volume is decreased so the pressure has to increase to keep the product of pressure and volume constant.
Complete step by step answer:
According to the question, the temperature of the air is constant in the process. So, the process of moving the piston is Isothermal. By ideal gas equation,
\[PV = nRT\]
The RHS of the equation as a whole is constant, so,
\[PV = const\]
Now when the air is moved to a smaller compartment from the big chamber to a smaller chamber the volume decreases. To keep the product constant, the pressure has to increase.
Option A and option D: There is no dependence of a gas’ root mean square velocity on volume so this is wrong.
Option B: This is not the correct explanation to the above situation. Whenever the gas is compressed there are more collisions of gas and walls and gas and gas particles. Due to collisions between gas particles the temperature increases. Here that is constant. So, the collisions between the gas and walls is the prominent reason as the mean free path for molecules reduces which leads to more frequent collisions.
Therefore, the correct answer is option C.
Note:Options B and C are confusing for most of us. Anywhere in kinetic theory of gases, there is no mention of the area of cross section in the development of the ideal gas equation. When a gas molecule hits the wall of the chamber, it requires very less area of that wall. So, the overall change in area of cross section is not prominent here. What is in fact important is that this overall reduced area leads to reduced mean free path lengths. This in turn causes more frequent collisions with the walls. The change in momentum w.r.t time is more which is nothing but increased force and hence increased pressure.
\[PV = const\]
So, the volume is decreased so the pressure has to increase to keep the product of pressure and volume constant.
Complete step by step answer:
According to the question, the temperature of the air is constant in the process. So, the process of moving the piston is Isothermal. By ideal gas equation,
\[PV = nRT\]
The RHS of the equation as a whole is constant, so,
\[PV = const\]
Now when the air is moved to a smaller compartment from the big chamber to a smaller chamber the volume decreases. To keep the product constant, the pressure has to increase.
Option A and option D: There is no dependence of a gas’ root mean square velocity on volume so this is wrong.
Option B: This is not the correct explanation to the above situation. Whenever the gas is compressed there are more collisions of gas and walls and gas and gas particles. Due to collisions between gas particles the temperature increases. Here that is constant. So, the collisions between the gas and walls is the prominent reason as the mean free path for molecules reduces which leads to more frequent collisions.
Therefore, the correct answer is option C.
Note:Options B and C are confusing for most of us. Anywhere in kinetic theory of gases, there is no mention of the area of cross section in the development of the ideal gas equation. When a gas molecule hits the wall of the chamber, it requires very less area of that wall. So, the overall change in area of cross section is not prominent here. What is in fact important is that this overall reduced area leads to reduced mean free path lengths. This in turn causes more frequent collisions with the walls. The change in momentum w.r.t time is more which is nothing but increased force and hence increased pressure.
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