Question:
$\text{The pressure-volume work for an ideal gas can be calculated by using the expression } W = \int_{V_i}^{V_f} P_{\text{ex}} \text{ d}V\text{.}$
$\text{The work can also be calculated from the pV-plot by using the area under the curve within the specified limits.}$
$\text{An ideal gas is compressed (a) reversibly or (b) irreversibly from volume } V_i \text{ to } V_f\text{.}$
$\text{The correct option is:}$
Solution:
$\text{Step 1: Understand the nature of reversible and irreversible processes}$
$\text{In a reversible process, the system changes infinitesimally slowly through a series of equilibrium states. The external pressure }(P_{\text{ex}})\text{ is only infinitesimally different from the gas pressure }(P)\text{, allowing the system to remain very close to equilibrium at all times.}$
$\text{In an irreversible process, the change happens rapidly. The external pressure is significantly different from the gas pressure, causing the system to be far from equilibrium during the transition.}$
$\text{Step 2: Analyze compression work for both processes}$
$\text{For compression (where }V_f < V_i\text{), work is done on the system, which is positive by convention:}$
$W = \int_{V_i}^{V_f} P_{\text{ex}} \text{ d}V$
$\text{Since }V_f < V_i\text{, the integral gives a negative value, but the work is positive (done on the system).}$
$\text{For a reversible compression, the external pressure is only slightly higher than the gas pressure, gradually increasing as the volume decreases. The P-V diagram shows a smooth curve.}$
$\text{For an irreversible compression, the external pressure is significantly higher and constant throughout the process. The P-V diagram shows a rectangular area.}$
$\text{Step 3: Compare the work done in both processes}$
$\text{When calculating the area under the curve (the work done), we find that:}$
$\text{For reversible compression: The work is represented by the area under the smooth curve from }V_i\text{ to }V_f\text{.}$
$\text{For irreversible compression: The work is represented by a rectangular area between }V_i\text{ and }V_f\text{ with a height of }P_{\text{ex}}\text{.}$
$\text{When comparing these areas for compression, the irreversible process requires more work than the reversible process:}$
$W \text{ (reversible)} < W \text{ (irreversible)}$
$\text{This is because in an irreversible process, the system is pushed away from equilibrium, resulting in energy dissipation and inefficiency. The excess work is lost as heat or other forms of energy dissipation.}$
$\text{For compression, more work is required for an irreversible process than for a reversible one to achieve the same final state.}$
$\text{Therefore, the correct option is 2: }W \text{ (reversible)} < W \text{ (irreversible)}$