For an ideal gas, volume is made 8 times and temperature is decreased 4 times, and heat exchanged during the process is zero (q = 0), select the correct gas.
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In adiabatic processes for ideal gases, the internal energy change is purely due to the change in temperature, and noble gases like He show this behavior perfectly.
Step 1: Using the First Law of Thermodynamics.
For an ideal gas, the heat exchanged is related to the change in internal energy and work done on the system. If \( q = 0 \), then the process is adiabatic. For an ideal gas, the internal energy depends only on temperature.
Step 2: Analyzing the gases.
- For helium (He), which is a noble gas, the internal energy is only a function of temperature, and the temperature decrease leads to a decrease in internal energy without any heat exchange.
- For other gases like CH$_4$, CO$_2$, and NH$_3$, their internal energies depend on both temperature and volume (since they have more complex interactions).
Step 3: Conclusion.
Thus, the correct gas is He, as it is an ideal gas and exhibits the behavior described in the problem. Hence, option (2) is the correct answer.
