MHTCET 16 May 2024 Shift 1 Paper

Concept:The kinetic theory of gases relates the pressure of a gas to the kinetic energy of its molecules per unit volume.Explanation:The pressure exerted by a gas is expressed as:$$P = \frac{1}{3} \rho v^{2}$$Here, $\rho$ is the density of the gas and $v$ is the root mean square speed of its molecules.Since density is mass per unit volume, we can write:$$P = \frac{1}{3} \left(\frac{M}{V}\right) v^{2}$$Multiply and divide the right-hand side by 2 to introduce kinetic energy:$$P = \frac{2}{3} \left(\frac{1}{2} \cdot \frac{M}{V} \cdot v^{2}\right)$$The kinetic energy per unit volume of the gas is:$$\text{K.E. per unit volume} = \frac{1}{2} \frac{M}{V} v^{2}$$Substituting this into the pressure equation gives:$$P = \frac{2}{3} \times \text{K.E. per unit volume}$$Thus, the pressure of the gas is equal to $\frac{2}{3}$ of its kinetic energy per unit volume.Answer:Pressure is $\left(\frac{2}{3}\right)^{\mathrm{rd}}$ of kinetic energy per unit volume of a gas. Hence, option (B) is correct.