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Question
The internal energy of an ideal gas decreases by the same amount as the work done by the system.
(a) The process must be adiabatic.
b) The process must be isothermal.
(c) The process must be isobaric.
(d) The temperature must decrease.
Solution
(a) The process must be adiabatic
(d) The temperature must decrease.
Using the first law of thermodynamics, we get
\[\Delta Q = \Delta W + \Delta U\]
\[ \Rightarrow \Delta Q = 0 .........\left[ \because \Delta W = - \Delta U \right]\]
Thus, no heat is exchanged in the process, i.e. the process is adiabatic and since the internal energy is decreasing, the temperature must decrease because the gas is an ideal gas.
On the other hand, volume and pressure of the gas are varying, leading to positive work done. So, the process cannot be isochoric and isobaric.
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