Question

The work done when one mole of an ideal gas expands at constant temperature $T$ from volume $V$ to $2V$ (in two equal steps of volume in a linear fashion) is $\frac{7}{12}RT$. How much more work would be done by the gas if it expands in three equal steps?
(R is the universal gas constant)}

• A. $\frac{1}{30}RT$
• B. $\frac{3}{8}RT$
• C. $\frac{3}{4}RT$
• D. $-RT \ln\left(\frac{1}{15}\right)$

Hint:

As the number of steps in an irreversible expansion increases, the process approaches reversible expansion, and the magnitude of work done by the gas increases.
The work for $n$ steps is given by the sum of series $\sum \frac{1}{n_i} RT$.

Answer 1

The Correct Option is A

Step 1 : Understanding the Question:
The question asks us to calculate the difference in work done by one mole of an ideal gas when it undergoes isothermal expansion from volume $V$ to $2V$ in three equal volume steps versus two equal volume steps.
Step 2 : Key Formulas and Approach:
For a step-wise, irreversible gas expansion, the work done in each step is:
\[ W = P_{\text{ext}} \Delta V \]
The external pressure ($P_{\text{ext}}$) for each step is equal to the final pressure of the gas at the end of that step.
For an ideal gas, $P = \frac{RT}{V_f}$.
Step 3 : Detailed Explanation:
Let us first verify the two-step expansion work $W_{(2)}$:
The total volume change is $2V - V = V$. For 2 equal steps, $\Delta V = \frac{V}{2} = 0.5V$.

Step 1: Expansion from $V$ to $1.5V$.
\[ P_{\text{ext}, 1} = P(1.5V) = \frac{RT}{1.5V} = \frac{2RT}{3V} \]
\[ W_1 = \left(\frac{2RT}{3V}\right)(0.5V) = \frac{1}{3}RT \]

Step 2: Expansion from $1.5V$ to $2V$.
\[ P_{\text{ext}, 2} = P(2V) = \frac{RT}{2V} \]
\[ W_2 = \left(\frac{RT}{2V}\right)(0.5V) = \frac{1}{4}RT \]

Total work:
\[ W_{(2)} = \left(\frac{1}{3} + \frac{1}{4}\right)RT = \frac{7}{12}RT \quad (\text{This matches the given value}) \]
Now, let us calculate the work done in three equal steps $W_{(3)}$:
Here, the volume change per step is $\Delta V = \frac{V}{3}$.
The volumes at the end of each step are $V_1 = \frac{4}{3}V$, $V_2 = \frac{5}{3}V$, and $V_3 = 2V$.

Step 1: Expansion from $V$ to $\frac{4}{3}V$.
\[ P_{\text{ext}, 1} = \frac{RT}{\frac{4}{3}V} = \frac{3RT}{4V} \implies W_1 = \left(\frac{3RT}{4V}\right)\left(\frac{V}{3}\right) = \frac{1}{4}RT \]

Step 2: Expansion from $\frac{4}{3}V$ to $\frac{5}{3}V$.
\[ P_{\text{ext}, 2} = \frac{RT}{\frac{5}{3}V} = \frac{3RT}{5V} \implies W_2 = \left(\frac{3RT}{5V}\right)\left(\frac{V}{3}\right) = \frac{1}{5}RT \]

Step 3: Expansion from $\frac{5}{3}V$ to $2V$.
\[ P_{\text{ext}, 3} = \frac{RT}{2V} \implies W_3 = \left(\frac{RT}{2V}\right)\left(\frac{V}{3}\right) = \frac{1}{6}RT \]

Total work:
\[ W_{(3)} = \left(\frac{1}{4} + \frac{1}{5} + \frac{1}{6}\right)RT = \left(\frac{15 + 12 + 10}{60}\right)RT = \frac{37}{60}RT \]
Let us calculate how much more work is done:
\[ \Delta W = W_{(3)} - W_{(2)} = \frac{37}{60}RT - \frac{7}{12}RT = \frac{37 - 35}{60}RT = \frac{2}{60}RT = \frac{1}{30}RT \]
Step 4 : Final Answer:
The additional work done in three steps is $\frac{1}{30}RT$.
This corresponds to Option (A).