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(D) Given condition: $\frac{P^2}{ho} = 	ext{constant}$.From the ideal gas equation,$P = \frac{ho RT}{M}$,so $ho = \frac{PM}{RT}$.Substituting $

Vedclass · Accepted Answer

Both $(B)$ and $(C)$.

Vedclass · Answer

(D) Given condition: $\frac{P^2}{ho} = 	ext{constant}$.From the ideal gas equation,$P = \frac{ho RT}{M}$,so $ho = \frac{PM}{RT}$.Substituting $ho$ into the given condition: $\frac{P^2}{PM/RT} = \frac{PRT}{M} = 	ext{constant}$.Since $R$ and $M$ are constants,$PT = 	ext{constant}$.This represents a rectangular hyperbola on a $P-T$ diagram,so statement $(C)$ is correct.Now,for the expansion where $ho_2 = ho/2$:Using $\frac{P_1^2}{ho_1} = \frac{P_2^2}{ho_2} \Rightarrow \frac{P^2}{ho} = \frac{P_2^2}{ho/2} \Rightarrow P_2^2 = \frac{P^2}{2} \Rightarrow P_2 = \frac{P}{\sqrt{2}}$.Using $P_1 T_1 = P_2 T_2$ (since $PT = 	ext{constant}$):$PT = \left(\frac{P}{\sqrt{2}}ight) T_2 \Rightarrow T_2 = \sqrt{2} T$.Thus,statement $(B)$ is also correct.Therefore,the correct option is $(D)$.

During an experiment,an ideal gas is found to obey a condition $\frac{P^2}{\rho} = \text{constant}$ [$\rho = \text{density of the gas}$]. The gas is initially at temperature $T$,pressure $P$,and density $\rho$. The gas expands such that density changes to $\rho/2$.

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