Question

Assuming the expression for the pressure exerted by the gas on the walls of the container, it can be shown the pressure is

& A.{{\left[ \dfrac{1}{3} \right]}^{^{rd}}}\text{kinetic energy per unit volume of a gas}\text{.} \\\ & B.{{\left[ \dfrac{2}{3} \right]}^{^{rd}}}\text{kinetic energy per unit volume of a gas}\text{.} \\\ & A.{{\left[ \dfrac{3}{3} \right]}^{^{rd}}}\text{kinetic energy per unit volume of a gas}\text{.} \\\ & \text{D}\text{.}\dfrac{3}{2}\times \text{kinetic energy per unit volume of a gas}\text{.} \\\ \end{aligned}$$

Answer 1

Here, we need to find the relationship between the pressure $P$ with respect to kinetic energy $K$ and volume of the gas $V$. We know from kinetic energy of the ideal gas $K=\dfrac{1}{2}mv^{2}$, where $v$ is the velocity, for gases $V_{rms}$ is taken, which is the root-mean square velocity, and $V_{rms}=\sqrt{\dfrac{3RT}{m}}$.

Formula used:
$K=\dfrac{1}{2}mv^{2}$ and $V_{rms}=\sqrt{\dfrac{3RT}{m}}$

Complete step by step answer:
We know that the kinetic energy of the ideal gas $K=\dfrac{1}{2}mv^{2}$, where $v$ is the velocity, for gases $V_{rms}$ is taken, which is the root-mean square velocity. And $m$ is the mass of the gas. We know that $V_{rms}=\sqrt{\dfrac{3kT}{ N_{a}m}}$ where $k$ is the Boltzmann constant, $T$ is the temperature in kelvin and $N_{a}$ is the Avogadro number.
Also, RMS of one mole of the gas molecules $V_{rms}=\sqrt{\dfrac{3RT}{m}}$, where $R$ is the molar gas constant.
Then replacing, we get, $K=\dfrac{1}{2}m\dfrac{3RT}{m}=\dfrac{3}{2} RT$
For one mole of ideal gas, $PV=RT$ , then $K=\dfrac{3}{2}PV$, where $P$ is the pressure exerted on the gases and $V$ is the volume of the gas.
Rearranging, we get $P=\dfrac{2K}{3V}$
Thus the answer is $B.\left[\dfrac{2}{3}\right]^{rd}\, kinetic\, energy\, per \,unit\, volume.$
Hence, the correct answer is option B.

Additional Information
$V_{rms}$ or the RMS velocity is the square root of the mean squares of the individual velocities of the gas molecules. It is given as$V_{rms}=\sqrt{\dfrac{3RT}{m}}$, where $R$ is the molar gas constant, $m$ is mass of the gas molecule and $T$ is the temperature in kelvin. It gives information on how fast or slow the molecules are moving in the average time. Since gas molecules undergo random motion, their mean velocity is taken for simplification.
Also, kinetic energy is the energy possessed by the molecules due to their motion in the container. It is given as $K=\dfrac{1}{2}mv^{2}$, in general. But for gases, $V_{rms}$ is taken instead of $v$.

Note:
To start with, write the formula of kinetic energy, and consider the same for ideal gases. $V_{rms}=\sqrt{\dfrac{3kT}{ N_{a}m}}$ where $k$ is the Boltzmann constant , $T$ is the temperature in kelvin and $N_{a}$ is the Avogadro number. For one mole of the gas molecules $V_{rms}=\sqrt{\dfrac{3RT}{m}}$, where $R$ is the molar gas constant. Only for one mole of ideal gas, $PV=RT$, is true.