The P-T diagram of an ideal gas having three | vedclass

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(B) For an ideal gas,the equation of state is given by:$PV = nRT$Since $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we have:$PV =

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$\rho_1 > \rho_2$

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(B) For an ideal gas,the equation of state is given by:$PV = nRT$Since $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we have:$PV = \frac{m}{M} RT$Rearranging for pressure $P$:$P = \left( \frac{m}{V} \right) \frac{RT}{M}$Since density $\rho = \frac{m}{V}$,we get:$P = \left( \frac{\rho R}{M} \right) T$This represents a straight line passing through the origin in a $P-T$ graph,where the slope is $\frac{\rho R}{M}$.For a constant temperature $T$,from the graph,we can see that $P_1 > P_2 > P_3$.Since $P \propto \rho$ for a constant $T$ and $M$,it follows that $\rho_1 > \rho_2 > \rho_3$.Therefore,the correct option is $\rho_1 > \rho_2$.

The $P-T$ diagram of an ideal gas having three different densities $\rho_1, \rho_2, \rho_3$ (in three different cases) is shown in the figure. Which of the following is correct?

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