Question 727: Thermodynamics & KTG

Question

Assuming the expression for the pressure exerted by the gas, it can be shown that pressure is

Accepted Answer

The pressure exerted by a gas on the walls of a container is derived from the kinetic theory of gases. It is represented by the equation:
$  P = \frac{1}{3} \rho v^2  $
where $ \rho $ is the density of the gas, and $ v $ is the root mean square (r.m.s.) speed of the gas molecules. We can further express this by noting that density ($\rho$) is mass ($M$) per unit volume ($V$):
$ P = \frac{1}{3} \left( \frac{M}{V} \right) v^2 $
Next, we multiply and divide the equation by $2$, to reframe it in terms of kinetic energy:
$ P = \frac{2}{3} \cdot \frac{1}{2} \left( \frac{M}{V} \right) v^2 $
Considering that the kinetic energy ($\mathrm{K.E.}$) of the gas per unit volume is given by:
$ \mathrm{K.E.} = \frac{1}{2} Mv^2 $
The expression for pressure can thus be rewritten as:
$ P = \frac{2}{3} \left( \frac{\mathrm{K.E.}}{V} \right) $
This equation shows that the pressure ($P$) is $\frac{2}{3}$ of the kinetic energy per unit volume of the gas.

Answer

$\left(\frac{3}{4}ight)^{	ext {th }}$ of kinetic energy per unit volume of a gas.

Answer

$\left(\frac{2}{3}ight)^{	ext {rd }}$ of kinetic energy per unit volume of a gas.

Answer

$\left(\frac{1}{3}\right)^{\mathrm{rd}}$ of kinetic energy per unit volume of a gas.

Answer

$\left(\frac{3}{2}ight)^{	ext {rd }}$ of kinetic energy per unit volume of a gas.

Thermodynamics & KTG

Assuming the expression for the pressure exerted by the gas, it can be shown that pressure is

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