When an ideal gas is taken from state a to b, along a path acb, 84 kJ of heat flows into the gas and the gas does 32 kJ of work. The following conclusions are drawn. Mark the one which is not correct.

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If the work done along the path adb is 10.5 kJ, the heat that will flow into the gas is 62.5 kJ.

When the gas is returned from b to a along the curved path, the work done on the gas is 21 kJ, and the system absorbs 73 kJ of heat.

lf Ua=0,Ud=42kJ, and the work done along the path adb is 10.5 kJ then the heat absorbed in the process ad is 52.5 kJ.

If Ua = 0, Ud = 42kJ, heat absorbed in the process db is 10 kJ.

answer is B.

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Detailed Solution

For path acb: Q=ΔU+W⇒ 84=ΔU+32⇒ΔU=52kJ Hence ΔUacb=ΔUab=ΔUadb=52kJFor path adb:Q=ΔU+W=52+10.5=62.5kJSo option (a) is correct.For process ba, system will release the heat. So option (b)is wrong.For path ad: Wadb=Wad+Wdb⇒ 10.5=ΔWad+0⇒ Wad=10.5kJ Qad=ΔUad+Wad=(42−0)+10.5=52.5kJSo option (c) is correct. Qadb=52+10.5=62.5kJQdb=Qadb−Qad=62.5−52.5=10kJSo option (d) is correct.Hence answer of this question is (b).

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