Subject is physical chemistry- ### Problem 5: Isothermal Monatomic Ideal Gas Expansion ProcessesConsider the following two isothermal monatomic ideal gas expansion processes:- **Path A**: Sudden (irreversible) decrease in pressure from \( P_1 \) to \( P_2 \) (where \( P_2 < P_1 \)).- **Path B**: Gradual (reversible) decrease in pressure from \( P_1 \) to \( P_2 \), such that the internal and external pressures remain in equilibrium at every step along the path.**Tasks:**(a) Draw a graph of \( P \) (pressure) vs. \( V \) (volume) illustrating the path followed in each of the above processes.(b) Obtain expressions for \(\Delta U\), \(W\), and \(Q\) for each of the above processes (express your results as functions of \( n \), \( T \), \( P_1 \), and/or \( P_2 \)).

Reversible Process: It is the process in which systems and surroundings can be changed back to their initial state from the final state without affecting the universe's thermodynamic properties. Irreversible Process: A process is specified to be irreversible when the system and surroundings do not return to their prior state, once the process has started. Given: In Path A, which is an irreversible process, the pressure P1 decreases to pressure P2 and P1 is greater than P2. In path B, in a reversible process, the pressure decreases from P1 to P2 but slowly.(a) A graph is plotted between pressure and volume to indicate path A and path B changes.(b) For reversible process: Calculation of W (work done): The work done (W), heat transferred (Q), and ∆U (internal energy change) has to be calculated. The formula to calculate work done (W) is as follows: dW =-PdV where dW is the small work done, P is the pressure and dV is the small change in volume. In the case of a reversible process, the external pressure P2 is equal to the internal pressure P1. The ideal gas equation is PV=nRT. Rearrange this equation for P as follows:P = nRTV Put this value of P in the expression to calculate the work done. dW =-nRTVdV Integrating on both sides with limits V1 and V2. ∫dW = -nRT∫V1V2dVV= -nRT [ln V]V2V1 = -nRT (ln V2 - ln V1)W =-nRT ln V2V1Calculation of ∆U: Now, ∆U is calculated. The internal energy change for an isothermal process is zero. Since in this reversible process, the temperature is constant, so it is an isothermal process and thus ∆U is zero. Calculation of Q: Q is calculated by subtracting the internal change from work done. The calculation is as follows: Q = ∆U - W= 0--nRT ln V2V1= nRT ln V2V1 For irreversible process: Calculation of W (work done): W = -Pext∆V=-P2 (V2 -V1)= -nRTV2(V2 -V1)= -nRT 1-V1V2 Calculation of ∆U: The internal energy change for an isothermal process is zero. Thus ∆U is zero. Calculation of Q: Q is calculated by subtracting the internal change from work done. The calculation is as follows: Q = ∆U - W= 0--nRT1-V1V2= nRT1-V1V2