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Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IIsothermal Process/Isobaric/IsochoricMedium2 minQB

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Passage / Comprehension

Passage: $Comprehension (Q.1 to Q.8) Use the thermodynamic equation of state (\frac{\partial U}{\partial V})_{T} = T (\frac{\partial P}{\partial T})_{V} - P to answer the following questions for n moles of a gas that obeys P = \frac{n RT}{V} + \frac{β n ^{2}}{V ^{2}} where β is a positive constant and C_{V} is independent of temperature.$

Heat absorbed by the gas in the isothermal change from ( $V_{1}$ , $T$ ) to ( $V_{2}$ , $T$ ) is

Options:

Answer:

Solution:

Work done by the gas: $\int P d V = n RT ln (V_{2} / V_{1}) - β n^{2} (1/ V_{2} - 1/ V_{1})$ .

$q = Δ U + w_{b y} = β n^{2} (1/ V_{2} - 1/ V_{1}) + n RT ln (V_{2} / V_{1}) - β n^{2} (1/ V_{2} - 1/ V_{1})$ .

The $β$ terms cancel: $q = n RT ln (V_{2} / V_{1})$ . Answer: (C).

Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:Isothermal Process/Isobaric/Isochoric

⏱ 2mℹ️ Source: QB

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