JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 7)
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 :
$${\left( 2 \right)^{{{1 + \gamma } \over 2}}}$$
2
$${\left( {{1 \over 2}} \right)^{{{1 + \gamma } \over 2}}}$$
$${\left( {{1 \over 2}} \right)^\gamma }$$
Explanation
$$\tau $$ $$ \propto $$ $${V \over {\sqrt T }}$$ ....(1)
Also we know, PV$$\gamma $$ = k
We know, PV = nRT
$$ \Rightarrow $$ P $$ \propto $$ $${T \over V}$$
$$ \therefore $$ $$\left( {{T \over V}} \right)$$V$$\gamma $$ = k
$$ \Rightarrow $$TV$$\gamma $$ - 1 = k
$$ \Rightarrow $$ T $$ \propto $$ V1 - $$\gamma $$
Using this value in equation (1)
$$\tau $$ $$ \propto $$ $${V \over {{V^{{{1 - \gamma } \over 2}}}}}$$
$$ \Rightarrow $$ $$\tau $$ $$ \propto $$ $${{V^{1 - {{1 - \gamma } \over 2}}}}$$
$$ \Rightarrow $$ $$\tau $$ $$ \propto $$ $${{V^{{{\gamma + 1} \over 2}}}}$$
$$ \therefore $$ $${{{\tau _2}} \over {{\tau _1}}}$$ = $${\left( {{{2V} \over V}} \right)^{^{{{\gamma + 1} \over 2}}}}$$ = $${\left( 2 \right)^{{{1 + \gamma } \over 2}}}$$
Also we know, PV$$\gamma $$ = k
We know, PV = nRT
$$ \Rightarrow $$ P $$ \propto $$ $${T \over V}$$
$$ \therefore $$ $$\left( {{T \over V}} \right)$$V$$\gamma $$ = k
$$ \Rightarrow $$TV$$\gamma $$ - 1 = k
$$ \Rightarrow $$ T $$ \propto $$ V1 - $$\gamma $$
Using this value in equation (1)
$$\tau $$ $$ \propto $$ $${V \over {{V^{{{1 - \gamma } \over 2}}}}}$$
$$ \Rightarrow $$ $$\tau $$ $$ \propto $$ $${{V^{1 - {{1 - \gamma } \over 2}}}}$$
$$ \Rightarrow $$ $$\tau $$ $$ \propto $$ $${{V^{{{\gamma + 1} \over 2}}}}$$
$$ \therefore $$ $${{{\tau _2}} \over {{\tau _1}}}$$ = $${\left( {{{2V} \over V}} \right)^{^{{{\gamma + 1} \over 2}}}}$$ = $${\left( 2 \right)^{{{1 + \gamma } \over 2}}}$$
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