JEE PHYSICS MCQ: THERMODYNAMICS

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Application +4 | -1

{C_V} = 4.96cal/mole K

,then increase in internal energy when temperature of 2 moles of this gas is increased from

340K

342K

View answer and explanation

Correct option: B

19.84cal

Option B (

19.84cal

) is correct because it's calculated using the formula for change in internal energy,

\Delta U = \mu C_V \Delta T

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