If the internal energy of an ideal gas decreases by the same amount as the work done by the system, then the

A Process must be isobaric
B Temperature must decrease
C Process must be adiabatic
D Both B and C

Correct Answer: Both B and C

By first law of thermodynamics we can say $$\delta Q = dU + \delta W$$
$$ \Rightarrow $$  for the internal energy and work to be same in magnitude we can say that heat transfer has to be zero $$\delta Q = 0$$
So the process is adiabatic and given the system is ideal gas and its internal energy is decreasing so, we can say that temperature decreases since the internal energy of a substance is solely dependent of temperature.

Related Questions

For an ideal gas, the internal energy depends upon its __________ only.
The internal energy of an ideal gas is a function of its __________ only.
Internal energy of an ideal gas

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