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.