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

Option: 1
the pressure of the gas changes to $\sqrt{2} \mathrm{P}$

Option: 2
the pressure of the gas changes to $\sqrt{2} \mathrm{~T}$

Option: 3
the graph of the above process on the $\mathrm{P-T}$ graph is parabola

Option: 4
the graph of the above process on the $\mathrm{P-T}$ graph is ellipse

Answers (1)

best_answer

$\mathrm{\frac{\mathrm{P}^2}{\rho}=\text { constant } }$

$\mathrm{ \Rightarrow \quad \frac{\mathrm{P}^2}{\rho}=\frac{\mathrm{P}_1^2}{\rho} \cdot 2}$

$\mathrm{\Rightarrow \mathrm{P}_1=\frac{\mathrm{P}}{\sqrt{2}} }$

$\mathrm{ \frac{\mathrm{P}^2}{\rho}=\text { constant } }$

$\mathrm{\Rightarrow \left.\frac{\rho^2 \mathrm{~T}^2}{\rho}=\text { constant } \quad \quad \text { [For ideal gas } \mathrm{P}=\rho \mathrm{RT}\right] }$

$\mathrm{\therefore \quad \rho \mathrm{T}^2=\text { constant } }$

$\mathrm{\therefore \quad \rho \mathrm{T}^2=\frac{\rho}{2} \mathrm{~T}_1^2 }$

$\mathrm{\Rightarrow \quad \mathrm{T}_1=\sqrt{2} \mathrm{~T} }$

$\mathrm{ \frac{\mathrm{P}^2}{\rho}=\text { constant } }$

$\mathrm{\Rightarrow \mathrm{P}^2 \frac{\mathrm{T}}{\mathrm{P}}=\text { constant } }$

$\mathrm{\therefore \text { PT = constant } }$

$\mathrm{\Rightarrow \text { Graph of process on } \mathrm{P}-\mathrm{T} \text { graph is hyperbola. } }$

Posted by

Rakesh

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