Question 410: Thermodynamics & KTG

Question

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 :

Accepted Answer

In adiabatic process
PV
$$\gamma $$
= constant
$$ \Rightarrow $$ $$P{\left( {{m \over ho }} ight)^\gamma }$$ = constant
As mass is constant
$$ 	herefore $$ P $$ \propto $$ $${{ho ^\gamma }}$$
$$ \Rightarrow $$ $${{{P_f}} \over {{P_i}}} = {\left( {{{{ho _f}} \over {{ho _i}}}} ight)^\gamma }$$ = $${\left( {32} ight)^{{7 \over 5}}}$$ = 2
7
= 128
[ For diatomic gas $$\gamma $$ = $${{7 \over 5}}$$ ]

Answer

128

Answer

32

Answer

326

Answer

$${1 \over {32}}$$

Thermodynamics & KTG

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 :

Choose the correct answer: