An ideal gas is taken through a process as shown in the figure. It absorbs 100 J of energy during the process AB. No heat is absorbed or rejected along the process BC and rejectes 140 J during the process CA. During the process BC, 60 J of work is done on the gas. Internal energy of the gas at A is 2000 J. The internal energy at C is

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1620 J

620 J

2160 J

216 J

answer is D.

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Detailed Solution

$AB:\Delta Q = \Delta U + \Delta W\$

+100 = U_B - U_A + zero

[Iso choric process work is zero]

$+ 100 = {U_B} - 2000 \Rightarrow {U_B} = 2100J$

$BC:\Delta Q = \Delta U + \Delta W$

O = (U_C - U_B) + (-60) [work is done on the gas]

$\Rightarrow {U_C} = {U_B} + 60 = 2100 + 60 = 2160J$

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