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  • A gas expands obeying the relation as shown in the diagram. The maximum temperature in th

A gas expands obeying the relation as shown in the \mathrm{P / V} diagram. The maximum temperature in this process is equal to


 

Option: 1

\mathrm{\frac{P_0 V_0}{R}}
 


Option: 2

\mathrm{\frac{3 P_0 V_0}{R}}
 


Option: 3

\mathrm{\frac{9 P_0 V_0}{8 R}}
 


Option: 4

None of these


Answers (1)

best_answer

The equation of \mathrm{P Q}  is given as

\mathrm{ \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1} }

\mathrm{ \Rightarrow \frac{P-P_0}{V-V_0}=\frac{P_0 / 2-P_0}{2 V_0-V_0} }

\mathrm{\Rightarrow \frac{P-P_0}{V-V_0}=-\frac{P_0}{2 V_0} }

\mathrm{ \Rightarrow 2 P V_0-2 P_0 V=-P_0 V+P_0 V_0 }

\mathrm{ \Rightarrow 2 P V_0+P_0 V=3 P_0 V }

\mathrm{ \Rightarrow P=\frac{3 P_0 V_0}{2 V_0}=\frac{3}{2} P_0-\frac{P_0}{2 V_0} V }

\mathrm{ \because P V=R T }

\mathrm{ \frac{R T}{V}=\frac{3}{2} P_0-\frac{P_0 V}{2 V_0} }

\mathrm{ T=\frac{3}{2} \frac{P_0 V}{R}-\frac{P_0 V^2}{2 V_0 R}}             ..........(1)

\mathrm{\frac{d T}{d V}=0 \ \frac{d^2 T}{d V^2}<0 \text { for } T_{\max } }

\mathrm{ \Rightarrow \frac{d T}{d V}=\frac{3 P_0}{2 R}-\frac{P_0 V}{V_0 R}=0 }

\mathrm{\Rightarrow V=\frac{3 V_0}{2}}

substituting this value of \mathrm{\mathrm{V}} in equation (1)

\mathrm{ \mathrm{T}_{\max }=\frac{9 \mathrm{P}_0 \mathrm{~V} 0}{8 \mathrm{R}} }

\therefore option 3 is correct.










 

Posted by

Divya Prakash Singh

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