Thermodynamics Question 260

Question: One mole of an ideal gas is taken from state A to state B by three different processes, (i) ACB (ii) ADB (iii) AEB as shown in the P-V diagram. The heat absorbed by the gas is-

Options:

A) greater in process (ii) than in (i)

B) the least in process (ii)

C) the same in (i) and (iii)

D) less in (iii) than in (ii)

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Answer:

Correct Answer: D

Solution:

[d] Heat absorbed by gas in three processes is given by $ Q _{ACB}=\Delta U+W _{ACB} $

$ Q _{ACB}=\Delta U $

$ Q _{ACB}=\Delta U+W _{AEB} $

The change in internal energy in all the three cases is same and $ W _{ACB} $ is $ +ve, $

$ W _{AEB} $ is$ -ve $ . Hence $ Q _{ACB}>Q _{ADB}>Q _{AEB} $



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