Question

A perfect gas (0.1 mol) having \( \bar{C}_V = 1.50 \) R (independent of temperature) undergoes the above transformation from point 1 to point 4. If each step is reversible, the total work done (w) while going from point 1 to point 4 is ____ J (nearest integer) [Given : R = 0.082 L atm K\(^{-1}\)] 

Hint:

For a cyclic process, the net work done is the area enclosed by the cycle on the P-V diagram. Work done during isobaric expansion is \( -P\Delta V \), and during isobaric compression is \( -P\Delta V \). Work done during isochoric processes is zero. Pay careful attention to the signs of work done during expansion and compression.

Answer 1

Correct Answer: 304

The problem asks for the calculation of the total work done (w) when a perfect gas undergoes a series of reversible transformations as depicted in the given P-V diagram, from an initial point 1 to a final point 4.

Concept Used:

The work done (w) in a reversible thermodynamic process is calculated by the integral of pressure with respect to volume. The convention used in chemistry is:

\[ w = - \int P_{ext} \, dV \]

For a reversible process, \(P_{ext} = P_{gas} = P\). The total work done for a process consisting of multiple steps is the sum of the work done in each individual step: \(w_{total} = w_{1 \to 2} + w_{2 \to 3} + w_{3 \to 4}\).

• For an isochoric process (constant volume), \(dV = 0\), so the work done is \(w = 0\).
• For an isobaric process (constant pressure), the work done is \(w = -P \Delta V = -P(V_{final} - V_{initial})\).

The final answer needs to be converted from L·atm to Joules (J) using the conversion factor \(1 \text{ L·atm} = 101.325 \text{ J}\).

Step-by-Step Solution:

Step 1: Analyze and calculate the work done for the path from point 1 to point 2 (\(w_{1 \to 2}\)).

This is an isochoric process, as the volume remains constant at \(V = 1000 \text{ cm}^3\). For any isochoric process, the change in volume \(dV\) is zero.

\[ \Delta V = V_2 - V_1 = 1000 - 1000 = 0 \text{ cm}^3 \] \[ w_{1 \to 2} = -P \Delta V = 0 \text{ J} \]

Step 2: Analyze and calculate the work done for the path from point 2 to point 3 (\(w_{2 \to 3}\)).

This is an isobaric process, as the pressure remains constant at \(P = 3.00 \text{ atm}\). The volume changes from \(V_2 = 1000 \text{ cm}^3\) to \(V_3 = 2000 \text{ cm}^3\).

First, calculate the change in volume \(\Delta V\):

\[ \Delta V = V_3 - V_2 = 2000 \text{ cm}^3 - 1000 \text{ cm}^3 = 1000 \text{ cm}^3 \]

Convert the volume from \(\text{cm}^3\) to Liters (L), since \(1 \text{ L} = 1000 \text{ cm}^3\):

\[ \Delta V = 1 \text{ L} \]

Now, calculate the work done in L·atm:

\[ w_{2 \to 3} = -P \Delta V = -(3.00 \text{ atm})(1 \text{ L}) = -3.00 \text{ Latm} \]

Step 3: Analyze and calculate the work done for the path from point 3 to point 4 (\(w_{3 \to 4}\)).

This is also an isochoric process, with the volume constant at \(V = 2000 \text{ cm}^3\).

\[ \Delta V = V_4 - V_3 = 2000 - 2000 = 0 \text{ cm}^3 \] \[ w_{3 \to 4} = -P \Delta V = 0 \text{ J} \]

Step 4: Calculate the total work done (\(w_{total}\)) for the entire transformation from 1 to 4.

The total work is the sum of the work done in each step.

\[ w_{total} = w_{1 \to 2} + w_{2 \to 3} + w_{3 \to 4} \] \[ w_{total} = 0 + (-3.00 \text{ Latm}) + 0 = -3.00 \text{ Latm} \]

Step 5: Convert the total work from L·atm to Joules (J).

Using the conversion factor \(1 \text{ Latm} = 101.325 \text{ J}\):

\[ w_{total} = -3.00 \text{ Latm} \times 101.325 \frac{\text{J}}{\text{Latm}} \] \[ w_{total} = -303.975 \text{ J} \]

Step 6: Round the result to the nearest integer as requested by the question.

\[ w_{total} \approx -304 \text{ J} \]

The question asks for the value to be filled in the blank of `(-)___ J`. Therefore, the value is the magnitude of the work done.

The total work done (w) is 304 J.