Thermodynamics Question 293
A gas is expanded from volume $ V _{0} $ to $ 2V _{0} $ under three different processes. Process 1 is isobaric, process 2 is isothermal, and process 3 is adiabatic. Let $ \Delta U _{1} $ , $ \Delta U _{2} $ and $ \Delta U _{3} $ be the change in internal energy of the gas in these three processes. Then-
Options:
A) $ \Delta U _{1}>\Delta U _{2}>\Delta U _{3} $
B) $ \Delta U _{1}<\Delta U _{2}<\Delta U _{3} $
C) $ \Delta U _{2}<\Delta U _{1}<\Delta U _{3} $
D) $ \Delta U _{2}<\Delta U _{3}<\Delta U _{1} $
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Answer:
Correct Answer: A
Solution:
Since volume is same in all three processes, therefore temperature will be least having least pressure.
BETA
