One mole of an ideal gas is taken from state A to state B by three different processes (a) ACB, (b) ADB and (c) AEB as shown in the P-V diagram. The heat absorbed by the gas is :
Answer & explanation
Correct answer: option 4
Heat absorbed by gas in three processes is given by :
\(Q_{ACB} = \Delta U + W_{ACB}\)
\(Q_{ADB} = \Delta U\)
\(Q_{AEB} = \Delta U + W_{AEB}\)
The change in internal energy in all the three cases is same and WACB is positive, WAEB is negative.
Hence, QACB > QADB > QAEB