One mol of an ideal gas was taken from A → B as shown in given figure. Magnitude of work involved in process is R = 25 3 J m o l K :

w = - n R Δ T = - 1 × 25 3 × 300 = - 2500 J

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ChemistryThermodynamics - 1Adiabatic Process/PolytropicEasy2 minPYQ_2020

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One mol of an ideal gas was taken from $A \to B$ as shown in given figure. Magnitude of work involved in process is $(R = \frac{25}{3} \frac{J}{m o l K})$ :

Options:

Answer:

Solution:

$w = - n R Δ T$

$= - 1 \times \frac{25}{3} \times 300$

$= - 2500 J$

Stream:NTA_ABHYASSubject:ChemistryTopic:Thermodynamics - 1Subtopic:Adiabatic Process/Polytropic

⏱ 2mℹ️ Source: PYQ_2020

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