Two different adiabatic paths for the same gas intersect two isot… - Vidyadip

MCQ

Two different adiabatic paths for the same gas intersect two isothermal curves as shown in$P-V$ diagram. The relation between the ratio $\frac{V_a}{V_d}$ and the ratio $\frac{V_b}{V_c}$ is:

A
$\frac{V_a}{V_d}=\left(\frac{V_b}{V_c}\right)^{-1}$
B
$\frac{V_a}{V_d} \neq \frac{V_b}{V_c}$
✓
$\frac{V_a}{V_d}=\frac{V_b}{V_c}$
D
$\frac{V_a}{V_d}=\left(\frac{V_b}{V_c}\right)^2$

✓

Answer

Correct option: C.

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

c
For adiabatic process

$\mathrm{TV}^{\gamma-1}=\text { constant }$

$\mathrm{T}_{\mathrm{a}} \cdot \mathrm{V}_{\mathrm{a}}^{\gamma-1}=\mathrm{T}_{\mathrm{d}} \cdot \mathrm{V}_{\mathrm{d}}^{\gamma-1}$

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

$\mathrm{T}_{\mathrm{b}} \cdot \mathrm{V}_{\mathrm{b}}^{\gamma-1}=\mathrm{T}_{\mathrm{c}} \cdot \mathrm{V}_{\mathrm{c}}^{\gamma-1}$

$\left(\frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{c}}}{\mathrm{T}_{\mathrm{b}}}$

$\mathrm{T}_{\mathrm{b}} \cdot \mathrm{V}_{\mathrm{b}}^{\gamma-1}=\mathrm{T}_{\mathrm{c}} \cdot \mathrm{V}_{\mathrm{c}}^{\gamma-1}$

$\left(\frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{c}}}{\mathrm{T}_{\mathrm{b}}}$

$\frac{\mathrm{V}_{\mathrm{a}}}{\mathrm{V}_{\mathrm{d}}}=\frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}} \quad\left(\begin{array}{r}\because \mathrm{T}_{\mathrm{d}}=\mathrm{T}_{\mathrm{c}} \\ \mathrm{T}_{\mathrm{a}}=\mathrm{T}_{\mathrm{b}}\end{array}\right)$

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