Question

The pressure p in terms of its mean kinetic energy/unit volume e is

a.e/3

b.e/2

c.3e/4

d.2e/3

e.5e/4

Answer 1

Answer: (d). The pressure is  $P= \dfrac{2e}{3}$.

Explanation:

Given that,

Kinetic energy /volume = e

Pressure = P

We know that,

$K.E = \dfrac{3 RT}{2}$.....(I)

Using ideal gas equation

$PV = RT$

Put the value of RT in equation (I)

$K.E = \dfrac{3 PV}{2}$

$\dfrac{K.E}{V} = \dfrac{3P}{2}$

$\dfrac{2K.E}{3 V} = P$

$P= \dfrac{2e}{3}$

Hence, the pressure is  $P= \dfrac{2e}{3}$

Answer 2

From the question, we need to express pressure in terms of e.

$e=\frac{Mean \quad kinetic \quad energy}{Unit \quad volume}$

Kinetic energy in terms of pressure and volume is given as,

$Kinetic\quad energy \quad (KE) = \frac {3}{2} P V$

$\Rightarrow \frac{\frac{2}{3} K E }{V} = P$

$\Rightarrow \frac{2 \times KE}{3 \times V} = P$

Thereby, in terms of mean kinetic energy per unit volume (e), we get,

$P = \frac {2}{3}e$