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Question

# Calculate q, w, ΔU for the isothermal reversible expansion of 1 mole of an ideal gas from an initial pressure of 1.0 bar to a final pressure of 0.1 bar at a constant temperature of 273 K.

A
w=q=ΔU=0
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B
w=q=5.22 kJ and ΔU=0
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C
q=w=5.22 kJ and ΔU=0
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D
q=w=6.22 kJ and ΔU=0
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Solution

## The correct option is C q=−w=5.22 kJ and ΔU=0In an isothermal process, temperature remains constant so ΔU is zero According to the first law of thermodynamics, ΔU=w+q0=w+qq=−wFor a reversible isothermal process,w=−2.303 nRT log10P1P2=−2.303×1×8.314×273log1010.1=−5227.1 J=−5.22 kJ Thus, q=−w=5.227 kJ and ΔU=0

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