Q.Explain, on the basis of kinetic theory, how the pressure of a gas changes if its volume is reduced at constant temperature.

[!TLDR] Reducing volume at constant temperature packs the same number of molecules closer, so they strike the walls more often, raising pressure. [!ANSWER] Reducing the volume at constant temperature increases the number density of molecules, so they collide with the walls more frequently, and the pressure increases -- consistent with P ∝ 1/V from kinetic theory.

From kinetic theory, the pressure of an ideal gas is P = (1/3)(N/V) m <v^2>, where N/V is the number of molecules per unit volume (the number density) and <v^2> is the mean square molecular speed. At constant temperature, the mean square speed <v^2> stays unchanged (since it depends only on T for a given gas, via <v^2> = 3kBT/m). If the volume V is reduced while the number of molecules N stays fixed, the number density N/V increases -- the same molecules are now confined to a smaller space. With more molecules in a given volume, molecules strike each unit area of the container walls more frequently in a given time, which directly increases the pressure. Since P is inversely proportional to V at fixed N and T (P ∝ 1/V, i.e. Boyle's law), reducing the volume increases the pressure in exact inverse proportion. [!ANSWER] Reducing the volume at constant temperature increases the number density of molecules, so they collide with the walls more frequently, and the pressure increases -- consistent with P ∝ 1/V from kinetic theory.

Start from P = (1/3)(N/V) m <v^2>, note <v^2> is fixed at constant T, and reason about how reducing V increases collision frequency with the walls.

Explaining the pressure increase only in terms of 'molecules being squeezed together' without connecting it to the actual mechanism -- increased collision FREQUENCY with the walls, not increased force per collision (which depends on speed, unchanged at constant T).