Which among the following is correct option for adiabatic reversible expansion of an ideal gas?
Let's analyze the characteristics of an adiabatic reversible expansion process involving an ideal gas. An adiabatic process is one where no heat is exchanged with the surroundings, meaning the heat flow, $\Delta Q$, is zero. A reversible process is an ideal process that can be reversed without leaving any change in the system or the surroundings. For an ideal gas, the internal energy depends only on temperature.
Consider the key thermodynamic quantities for this process:
$\Delta Q = 0$
$\Delta S_{surr} = \frac{Q_{surr}}{T_{surr}}$
In an adiabatic process, $Q_{system} = 0$. By the first law of thermodynamics, $Q_{surr} = -Q_{system} = 0$. Therefore, the entropy change for the surroundings is zero.$\Delta S_{surr} = 0$
$\Delta S_{total} = 0$
Combined with $\Delta S_{surr} = 0$, this implies that the entropy change for the system ($\Delta S_{system}$) is also zero for a reversible adiabatic process.$\Delta S_{system} = 0$
$\Delta T < 0$
This confirms that the temperature change is not zero.$\Delta T \ne 0$
Based on our analysis:
Let's check the provided options against these findings. The option that states 'Entropy change for surroundings = 0, Total entropy change = 0, $\Delta T \ne 0$' matches our derived characteristics for an adiabatic reversible expansion of an ideal gas.
When a gas is compressed suddenly then its temperature
During throttling process:
2 mol of a monoatomic ideal gas with initial volume of 5 L and pressure 10 bar undergoes an irreversible adiabatic expansion against a constant final pressure of 1 bar. The final volume (in L) is ________.
(Given: R = $8.314 \times 10^{-2}$ L bar $mol^{-1}$ $K^{-1}$)
(rounded off to one decimal place)

From the above Carnot cycle undergone by an ideal gas, identify the processes in which the change in internal energy is NON-ZERO.