An adiabatic vessel of total volume V is partitioned into two equal parts by a fixed, conducting separator. The left part holds one mole of an ideal gas ( U = 1.5 nRT ) and the right part contains two moles of the same

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Correct answer: D. (pV)/(2R) and (pV)/(4R)

Total volume of the vessel is V . Since the vessel is partitioned into two equal parts, the volume of each part is (V)/(2) . For the left part, using the ideal gas equation pV = nRT : p ((V)/(2)) = 1 × R × T_ L T_ L = (pV)/(2R) For the right part, using the ideal gas equation pV = nRT : p ((V)/(2)) = 2 × R × T_ R T_ R = (pV)/(4R) The initial temperatures on the left and right sides are (pV)/(2R) and (pV)/(4R) respectively.

An adiabatic vessel of total volume V is partitioned into two equal...

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