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NCERT Exemplar · Q13

Q.When an ideal gas is compressed adiabatically, its temperature rises: the molecules on the average have more kinetic energy than before. The kinetic energy increases, (Note: more than one of the given options may be correct.)

(a) because of collisions with moving parts of the wall only.
(b) because of collisions with the entire wall.
(c) because the molecules gets accelerated in their motion inside the volume.
(d) because of redistribution of energy amongst the molecules.
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During adiabatic compression, molecules gain kinetic energy exclusively through collisions with the moving inward wall (the piston), not from the stationary walls or internal redistribution. The correct option is (A).

Kinetic Theory Explanation

When a gas is compressed adiabatically, no heat flows in or out (Q=0Q = 0). Yet the temperature rises, meaning the average molecular kinetic energy increases. Where does this energy come from?

The answer lies in understanding what happens at the microscopic level during compression. A gas molecule bouncing off a surface behaves like an elastic collision. If the surface is stationary, the molecule rebounds with the same speed it arrived with—no energy change. But if the surface is moving toward the molecule, the rebound speed is higher than the incident speed, and the molecule gains kinetic energy. This is exactly what happens to a tennis ball bouncing off an approaching racket.

In adiabatic compression, the piston (the moving part of the wall) advances into the gas. Molecules colliding with this moving piston pick up extra kinetic energy with each bounce. The stationary walls contribute nothing to the energy increase—they merely reflect molecules elastically. Internal collisions between molecules conserve total kinetic energy and only redistribute it; they cannot create new energy.

Let's examine each option:

Step-by-Step Analysis

  1. Option (A): Collisions with moving parts of the wall only

    Consider a molecule of mass mm approaching the piston with velocity component vxv_x perpendicular to the piston. If the piston moves inward with speed uu, the molecule's velocity in the piston's reference frame is (vx+u)(v_x + u). After elastic reflection, it rebounds with velocity (vx+u)(v_x + u) in the opposite direction relative to the piston. Transforming back to the lab frame, the molecule's velocity after collision is (vx+2u)(v_x + 2u).

    The kinetic energy change is:

ΔKE=12m(vx+2u)2−12mvx2=2muvx+2mu2\Delta KE = \frac{1}{2}m(v_x + 2u)^2 - \frac{1}{2}mv_x^2 = 2muv_x + 2mu^2

For typical molecular speeds much larger than piston speed (vx≫uv_x \gg u), the dominant term is 2muvx>02muv_x > 0. Each collision with the moving piston increases molecular kinetic energy. This is the only mechanism by which work done on the gas converts to internal energy.

  1. Option (B): Collisions with the entire wall

    The "entire wall" includes both the moving piston and the stationary container walls. Stationary walls reflect molecules elastically with no change in kinetic energy—the velocity component perpendicular to the wall reverses, but the speed remains constant. Only the moving part (the piston) contributes to the energy increase. This option is incorrect because it attributes the energy gain to all walls indiscriminately.

  2. Option (C): Molecules get accelerated in their motion inside the volume …

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