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Two different adiabatic paths for the same gas intersect two isothermals at \mathrm{T_1 \text { and } T_2}as shown in the P −V diagram (Fig). Then

Option: 1

\mathrm{\frac{V_a}{V_c}=\frac{V_b}{V_d}}


Option: 2

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


Option: 3

\mathrm{\frac{V_a}{V_b}=\frac{V_d}{V_c}}


Option: 4

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


Answers (1)

best_answer

The two adiabatic paths ad and bc for the gas intersect the two isothermals ab and cd
at temperatures \mathrm{T_1 \text { and } T_2} (see Fig.). Since points a and d lie on the same adiabatic path,
we have

\mathrm{\begin{aligned} & T_1 V_a^{(\gamma-1)}=T_2 V_d^{(\gamma-1)} \\ & \operatorname{Or}\left(\frac{V_a}{V_d}\right)^{(\gamma-1)}=\frac{T_2}{T_1} \end{aligned}}                   [1]

Since points b and c also lie on the same adiabatic path,

\mathrm{\begin{aligned} & T_1\left(V_b\right)^{(\gamma-1)}=T_2 V_c^{(\gamma-1)} \\ & \operatorname{Or}\left(\frac{V_b}{V_c}\right)^{(\gamma-1)}=\frac{T_2}{T_1} \end{aligned}}                [2]

From (1) and (2), we get

\mathrm{\begin{aligned} & \left(\frac{V_a}{V_d}\right)^{(\gamma-1)}=\left(\frac{V_b}{V_c}\right)^{(\gamma-1)} \\ & \operatorname{Or} \frac{V_a}{V_d}=\frac{V_b}{V_c} \end{aligned}}

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Shailly goel

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