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  • Two adiabatic processes for the same gas are given to interse

Two adiabatic processes \mathrm{bc\: and \: ad} for the same gas are given to intersect two isotherms at \mathrm{T_1\: and\: T_2} (as shown). Then


 

Option: 1

\mathrm{\frac{V_a}{V_b}=\frac{T_2}{T_1}}
 


Option: 2

\mathrm{\frac{V_a}{V_d}=\frac{T_1}{T_2}}
 


Option: 3

\mathrm{\frac{V_b}{V_c}=\left(\frac{T_2}{T_1}\right)^{\frac{1}{\gamma -1}}}
 


Option: 4

\mathrm{V_a V_c \neq V_b V_d}


Answers (1)

best_answer

For,\mathrm{A,D}

\mathrm{ \mathrm{T}_1 \mathrm{~V}_{\mathrm{a}}^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_{\mathrm{d}}^{\gamma-1} }

\mathrm{ \left(\frac{\mathrm{V}_{\mathrm{a}}}{\mathrm{V}_{\mathrm{d}}}\right)^{\gamma-1}=\frac{\mathrm{T}_2}{\mathrm{~T}_1} }

For \mathrm{B, C}

\mathrm{ \mathrm{T}_1 \mathrm{~V}_{\mathrm{b}}^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_{\mathrm{c}}^{\gamma-1} }

\mathrm{ \left(\frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}}\right)^{\gamma-1}=\frac{\mathrm{T}_2}{\mathrm{~T}_1} \Rightarrow \frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}}=\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^{\frac{1}{\gamma-1}}}

So \mathrm{\left(\frac{V_a}{V_d}\right)=\left(\frac{V_b}{V_c}\right)}, i.e., choices C iscorrect. Hence choice A, B , D are wrong.

Posted by

Ajit Kumar Dubey

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