JEE MAIN - Physics (2020 - 5th September Evening Slot - No. 8)
In an adiabatic process, the density of a
diatomic gas becomes 32 times its initial value.
The final pressure of the gas is found to be n
times the initial pressure. The value of n is :
128
32
326
$${1 \over {32}}$$
Explanation
In adiabatic process
PV$$\gamma $$ = constant
$$ \Rightarrow $$ $$P{\left( {{m \over \rho }} \right)^\gamma }$$ = constant
As mass is constant
$$ \therefore $$ P $$ \propto $$ $${{\rho ^\gamma }}$$
$$ \Rightarrow $$ $${{{P_f}} \over {{P_i}}} = {\left( {{{{\rho _f}} \over {{\rho _i}}}} \right)^\gamma }$$ = $${\left( {32} \right)^{{7 \over 5}}}$$ = 27 = 128
[ For diatomic gas $$\gamma $$ = $${{7 \over 5}}$$ ]
PV$$\gamma $$ = constant
$$ \Rightarrow $$ $$P{\left( {{m \over \rho }} \right)^\gamma }$$ = constant
As mass is constant
$$ \therefore $$ P $$ \propto $$ $${{\rho ^\gamma }}$$
$$ \Rightarrow $$ $${{{P_f}} \over {{P_i}}} = {\left( {{{{\rho _f}} \over {{\rho _i}}}} \right)^\gamma }$$ = $${\left( {32} \right)^{{7 \over 5}}}$$ = 27 = 128
[ For diatomic gas $$\gamma $$ = $${{7 \over 5}}$$ ]
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