Question 510: Thermodynamics & KTG

Question

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as $${V^q},$$ where $$V$$ is the volume of the gas. The value of $$q$$ is: $$\left( {\gamma  = {{{C_p}} \over {{C_v}}}} \right)$$

Accepted Answer

$$	au  = {1 \over {\sqrt 2 \pi {d^2}\left( {{N \over V}} ight)\sqrt {{{3RT} \over M}} }}$$
$$	au  \propto {V \over {\sqrt T }}$$
As, $$\,\,\,\,T{V^{\gamma  - 1}} = K$$
So, $$\,\,\,\,	au  \propto {V^{\gamma  + 1/2}}$$
Therefore, $$q = {{\gamma  + 1} \over 2}$$

Answer

$${{\gamma  + 1} \over 2}$$

Answer

$${{\gamma  - 1} \over 2}$$

Answer

$${{3\gamma  + 5} \over 6}$$

Answer

$${{3\gamma  - 5} \over 6}$$

Thermodynamics & KTG

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as ${V^q}, where V is the volume of the gas. The value of q is: \left( {\gamma = {{{C_p}} \over {{C_v}}}} \right)

Choose the correct answer: