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Chemistry - Real Gas Question with Solution | TestHub

ChemistryReal GasBoyle TemperatureHard2 minQB

ChemistryHardinteger

A van der Waals gas has a critical pressure of 41 atm and a critical volume of $0.15 L / mol$ . Its Boyle temperature, in kelvin, is $(R = 0.082 L atm K^{- 1} mol^{- 1})$

Answer:

675

Solution:

$V_{c} = 3 b$ , so $b = 0.15/3 = 0.05 L / mol$ .

$P_{c} = a / (27 b^{2})$ and $V_{c} = 3 b$ give $a = 3 V_{c}^{2} P_{c} = 3 \times 0.0225 \times 41 = 2.7675 atm L^{2} / mol^{2}$ .

$T_{B} = a / (R b) = 2.7675/ (0.082 \times 0.05) = 675 K$ .

Answer: 675.

Stream:JEESubject:ChemistryTopic:Real GasSubtopic:Boyle Temperature

⏱ 2mℹ️ Source: QB

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