JEE PYQ: Thermodynamics Question 23
Question 23 - 2020 (05 Sep Shift 2)
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 :
(1) 32
(2) 326
(3) 128
(4) $\frac{1}{32}$
Show Answer
Answer: (3)
Solution
In adiabatic process $PV^{\gamma} =$ constant
$\therefore P\left(\frac{m}{\rho}\right)^{\gamma} =$ constant $\left(\because V = \frac{m}{\rho}\right)$
As mass is constant, $\therefore P \propto \rho^{\gamma}$
If $P_i$ and $P_f$ be the initial and final pressure of the gas and $\rho_i$ and $\rho_f$ be the initial and final density of the gas. Then
$\frac{P_f}{P_i} = \left(\frac{\rho_f}{\rho_i}\right)^{\gamma} = (32)^{7/5}$
$\frac{nP_i}{P_i} = (2^5)^{7/5} = 2^7$
$\Rightarrow n = 2^7 = 128$
BETA
