Question 433: Thermodynamics & KTG

Question

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean
collision time between the gas molecule changes from $${	au _1}$$
 to $${	au _2}$$
. If $${{{C_p}} \over {{C_v}}} = \gamma $$ for this gas then a good
estimate for $${{{	au _2}} \over {{	au _1}}}$$
 is given by :

Accepted Answer

$$	au $$ $$ \propto $$ $${V \over {\sqrt T }}$$ ....(1)
Also we know, PV
$$\gamma $$
= k
We know, PV = nRT
$$ \Rightarrow $$ P $$ \propto $$ $${T \over V}$$
$$ 	herefore $$ $$\left( {{T \over V}} ight)$$V
$$\gamma $$
= k
$$ \Rightarrow $$TV
$$\gamma $$ - 1
= k
$$ \Rightarrow $$ T $$ \propto $$ V
1 - $$\gamma $$
Using this value in equation (1)
$$	au $$ $$ \propto $$ $${V \over {{V^{{{1 - \gamma } \over 2}}}}}$$
$$ \Rightarrow $$ $$	au $$ $$ \propto $$ $${{V^{1 - {{1 - \gamma } \over 2}}}}$$
$$ \Rightarrow $$ $$	au $$ $$ \propto $$ $${{V^{{{\gamma  + 1} \over 2}}}}$$
$$ 	herefore $$ $${{{	au _2}} \over {{	au _1}}}$$ = $${\left( {{{2V} \over V}} ight)^{^{{{\gamma  + 1} \over 2}}}}$$ = $${\left( 2 ight)^{{{1 + \gamma } \over 2}}}$$

Answer

$${\left( 2 \right)^{{{1 + \gamma } \over 2}}}$$

Answer

2

Answer

$${\left( {{1 \over 2}} \right)^{{{1 + \gamma } \over 2}}}$$

Answer

$${\left( {{1 \over 2}} \right)^\gamma }$$

Thermodynamics & KTG

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from ${\tau _1} to {\tau _2} . If {{{C_p}} \over {{C_v}}} = \gamma for this gas then a good estimate for {{{\tau _2}} \over {{\tau _1}}}$ is given by :

Choose the correct answer: