Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IFirst law of ThermodynamicsMedium2 minPYQ_2021

ChemistryMediumsingle choice

Work out the heat change (cal) when $40 g$ of He gas at $2 7^{\circ} C$ undergoes isothermal and reversible compression from initial pressure of $1 atm$ to $10 atm (R = 2 cal K^{- 1} mol^{- 1})$ .

Options:

Answer:

Solution:

$q = - w_{rev} = - (- 2.303 nRT lo g \frac{P _{1}}{P _{2}}) = 2.303 \times \frac{40}{4} \times 2 \times 300 lo g \frac{1}{10} = - 13.82 kcal$

Stream:BITSATSubject:ChemistryTopic:Thermodynamics - ISubtopic:First law of Thermodynamics

⏱ 2mℹ️ Source: PYQ_2021

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