Q167 / 200 · MHT CET - Medical 2026 (24 April Shift 2) · Physics
Pressure exerted by an ideal gas on the walls of the container is equal to
a) mean kinetic energy per unit volume.
b) half of mean kinetic energy per unit volume.
c) one-third of mean kinetic energy per unit volume.
d) two-third of mean kinetic energy per unit volume.
a) mean kinetic energy per unit volume.
b) half of mean kinetic energy per unit volume.
c) one-third of mean kinetic energy per unit volume.
d) two-third of mean kinetic energy per unit volume.
- Correct answerAd
- Bc
- Cb
- Da
Solution
The pressure exerted by an ideal gas is given by $P = \dfrac{1}{3} \rho v_{rms}^2$, where $\rho$ is the density of the gas and $v_{rms}$ is the root mean square velocity.
The mean kinetic energy per unit volume is given by $E = \dfrac{1}{2} \rho v_{rms}^2$.
From these two equations, we can write $P = \dfrac{2}{3} \left( \dfrac{1}{2} \rho v_{rms}^2 \right) = \dfrac{2}{3} E$.
Therefore, the pressure exerted by an ideal gas is equal to two-thirds of the mean kinetic energy per unit volume.
Answer: d
The mean kinetic energy per unit volume is given by $E = \dfrac{1}{2} \rho v_{rms}^2$.
From these two equations, we can write $P = \dfrac{2}{3} \left( \dfrac{1}{2} \rho v_{rms}^2 \right) = \dfrac{2}{3} E$.
Therefore, the pressure exerted by an ideal gas is equal to two-thirds of the mean kinetic energy per unit volume.
Answer: d
Hence, the correct answer is option: 'A'.