Which of the following graphs correctly represents the variation of β = − 1 V dV dP with P for an ideal gas at constant temperature?

PV = Constant On differentiating both sides, we get PdV + VdP = 0 PdV = -VdP d V d P = - V P 1 P = − 1 V dV dP = β Since β = 1 P ; inverse relation result into rectangular hyperbola.

Chemistry - Ideal Gas Question with Solution | TestHub

ChemistryIdeal GasMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasysingle choice

Which of the following graphs correctly represents the variation of $β = - \frac{1}{V} \frac{dV}{dP}$ with P for an ideal gas at constant temperature?

Options:

Answer:

Solution:

PV = Constant
On differentiating both sides, we get
PdV + VdP = 0
PdV = -VdP
$\frac{d V}{d P} = - \frac{V}{P}$
$\frac{1}{P} = - \frac{1}{V} \frac{dV}{dP} = β$
Since $β = \frac{1}{P}$ ; inverse relation result into rectangular hyperbola.

Stream:NTA_ABHYASSubject:ChemistryTopic:Ideal GasSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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