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Integration Using Substitution

The P-V dia...

Question

The $P - V$ diagram shows that two adiabatic parts of the same gas intersect two isothermals at $T_{1}$ and $T_{2}$ . How the ratio $(V_{a} / V_{d})$ and $(V_{b} / V_{c})$ are related to each other?

(\frac{V_{a}}{V_{d}}) = 2 (\frac{V_{b}}{V_{c}})

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(\frac{V_{a}}{V_{d}}) = (\frac{V_{b}}{V_{c}})

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(\frac{V_{a}}{V_{d}}) = (\frac{V_{b}}{V_{d}})

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(\frac{V_{a}}{V_{c}}) = {(\frac{V_{b}}{V_{c}})}^{2}

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Solution

The correct option is A $(\frac{V_{a}}{V_{d}}) = (\frac{V_{b}}{V_{c}})$
For adiabatic process $P V^{γ} =$ constant and
$P V = n R T$
$∴ T V^{γ - 1} =$ constant
BC is adiabatic, $T_{1} V_{b}^{γ - 1} = T_{2} V_{c}^{γ - 1}$
AD is adiabatic, $T_{1} V_{a}^{γ - 1} = T_{2} V_{d}^{γ - 1}$
So, ${(\frac{V_{a}}{V_{b}})}^{γ - 1} = {(\frac{V_{d}}{V_{c}})}^{γ - 1}$

$\Rightarrow \frac{V_{a}}{V_{d}} = \frac{V_{b}}{V_{c}}$ .

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The P−V diagram shows that two adiabatic parts of the same gas intersect two isothermals at T1 and T2. How the ratio (Va/Vd) and (Vb/Vc) are related to each other?

The correct option is A (VaVd)=(VbVc)For adiabatic process PVγ= constant and PV=nRT∴TVγ−1=constantBC is adiabatic, T1Vγ−1b=T2Vγ−1cAD is adiabatic, T1Vγ−1a=T2Vγ−1dSo, (VaVb)γ−1=(VdVc)γ−1⇒VaVd=VbVc.

The $P - V$ diagram shows that two adiabatic parts of the same gas intersect two isothermals at $T_{1}$ and $T_{2}$ . How the ratio $(V_{a} / V_{d})$ and $(V_{b} / V_{c})$ are related to each other?