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} $