Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IFirst law of ThermodynamicsMedium2 minQB

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The average degree of freedom per molecule for a gas is 5. The gas absorbs 140 J of heat when it expands at constant pressure. Find the work done by the gas.

Options:

Answer:

Solution:

$C_{V} = \frac{f}{2} R = 2.5 R$ , so $C_{P} = 3.5 R$ .

At constant pressure $q = n C_{P} Δ T = 3.5 n R Δ T$ and $w = n R Δ T$ .

$w = \frac{q}{3.5} = \frac{140}{3.5} = 40$ J.

(100 J is $Δ U = q - w$ .)

Answer: (B).

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

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