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Thermodynamics question

2017 · Q109
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Thermodynamics question

2017 · Q109

NEETChemistryThermodynamicsMCQ+4 / −1
A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy Δ\DeltaΔU of the gas in joules will be
  1. A
    −-− 500 J
  2. B
    −-− 505 J
  3. C
    + 505 J
  4. D
    1136.25 J
View written solutionFree

Correct answer: B

w = - PextΔ\Delta ΔV = -2.5(4.50 - 2.50)

⇒\Rightarrow⇒ - 5 L atm = - 5 ×\times× 1.01.325 J = - 506.625 J

Δ\Delta ΔU = q + w

As, the container is insulted, thus q = 0

Hence, Δ\Delta ΔU = w = -506.625 J

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