An ideal gas expands in volume from 1 × 1 0 − 3 m 3 to 1 × 1 0 − 2 m 3 at 300 K against a constant pressure of 1 × 1 0 5 Nm − 2 . The work done is

Work done = − P ( Δ V ) = − 1 × 1 0 5 [ 1 0 − 2 − 1 0 − 3 ] = − 900 J

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ChemistryThermodynamics - IFirst law of ThermodynamicsMedium2 minPYQ_2004

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An ideal gas expands in volume from $1 \times 1 0^{- 3} m^{3}$ to $1 \times 1 0^{- 2} m^{3}$ at $300 K$ against a constant pressure of $1 \times 1 0^{5} Nm^{- 2}$ . The work done is

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Answer:

Solution:

Work done $= - P (Δ V)$ $= - 1 \times 1 0^{5} [1 0^{- 2} - 1 0^{- 3}] = - 900 J$

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

⏱ 2mℹ️ Source: PYQ_2004

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