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Physics
Pressure versus temperature graph of an ideal gas is as shown in figure. Density of the gas at point A is ρ0. Density at point B will be
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Q. Pressure versus temperature graph of an ideal gas is as shown in figure. Density of the gas at point $A$ is $\rho_{0}$. Density at point $B$ will be
AIIMS
AIIMS 2010
Kinetic Theory
A
$\frac{3}{4}\rho_{0}$
0%
B
$\frac{3}{2}\rho_{0}$
54%
C
$\frac{4}{3}\rho_{0}$
31%
D
$2\rho_{0}$
15%
Solution:
$ \rho=\frac{P M}{R T}$ or $\rho \propto \frac{P}{T}$
$\left(\frac{P}{T}\right)_{A}=\frac{P_{0}}{T_{0}} $ and $\left(\frac{P}{T}\right)_{ B }=\frac{3}{2}\left(\frac{P_{0}}{T_{0}}\right)$
$\left(\frac{P}{T}\right)_{B}=\frac{3}{2}\left(\frac{P}{T}\right)_{A}$
$\therefore \rho_{B}=\frac{3}{2} \rho_{A}=\frac{3}{2} \rho_{0}$