An ideal gas undergoes a cyclic transformation starting from the point \( A \) and coming back to the same point by tracing the path \( A \to B \to C \to A \) as shown in the diagram. The total work done in the process is \( \_\_\_\_ \) J.
The problem requires calculating the total work done by an ideal gas undergoing a cyclic process A → B → C → A as shown in the given V-P diagram.
Concept Used:
The total work done in a cyclic thermodynamic process is the sum of the work done in each individual step of the cycle. The work done by a gas during a volume change from an initial state to a final state is given by the integral:
\[ W = \int_{V_i}^{V_f} P \, dV \]
For specific processes:
Isobaric process (constant pressure): \( W = P \Delta V = P(V_f - V_i) \)
Isochoric process (constant volume): \( W = 0 \) since \( dV = 0 \).
The net work done in a cycle can also be found from the area enclosed by the cycle on a P-V diagram. However, here we will calculate the work for each step individually from the given V-P diagram and sum them up.
The required unit conversion is: \( 1 \text{ kPa} \times 1 \text{ dm}^3 = (10^3 \text{ Pa}) \times (10^{-3} \text{ m}^3) = 1 \text{ J} \).
Step-by-Step Solution:
Step 1: Identify the coordinates (Pressure, Volume) for points A, B, and C from the given V-P diagram.
The y-axis represents Volume (V) in dm³ and the x-axis represents Pressure (P) in kPa.
The total work done in the cycle is \( W_{total} = W_{A \to B} + W_{B \to C} + W_{C \to A} \).
Step 2: Calculate the work done for the path A → B.
This is an isobaric process since the pressure is constant at \( P_A = P_B = 10 \) kPa. The volume increases from 10 dm³ to 30 dm³, so this is an expansion.
Step 3: Calculate the work done for the path C → A.
This is an isochoric process since the volume is constant at \( V_C = V_A = 10 \) dm³. For any isochoric process, the work done is zero.
\[ W_{C \to A} = 0 \text{ J} \]
Step 4: Calculate the work done for the path B → C.
In this process, both pressure and volume change. The path is a straight line on the V-P diagram connecting B(10 kPa, 30 dm³) and C(30 kPa, 10 dm³). The work done is the area under this line on a P-V diagram. Since the volume decreases from 30 dm³ to 10 dm³, this is a compression, and the work done will be negative.
The area under the line segment BC on a P-V diagram is a trapezoid. The magnitude of the work is the area of this trapezoid.
Enthalpy Change refers to the difference between the heat content of the initial and final state of the reaction. Change in enthalpy can prove to be of great importance to find whether the reaction is exothermic or endothermic.
Formula for change in enthalpy is:-
dH = dU + d(PV)
The above equation can be written in the terms of initial and final states of the system which is defined below:
UF – UI = qP –p(VF – VI)
Or qP = (UF + pVF) – (UI + pVI)
Enthalpy (H) can be written as H= U + PV. Putting the value in the above equation, we obtained:
qP = HF – HI = ∆H
Hence, change in enthalpy ∆H = qP, referred to as the heat consumed at a constant pressure by the system. At constant pressure, we can also write,
∆H = ∆U + p∆V
Standard Enthalpy of Reaction
To specify the standard enthalpy of any reaction, it is calculated when all the components participating in the reaction i.e., the reactants and the products are in their standard form. Therefore the standard enthalpy of reaction is the enthalpy change that occurs in a system when a matter is transformed by a chemical reaction under standard conditions.
