Ratio of Work Done Along Two Paths for Ideal Gas

To find the ratio of the work done along the dashed line path ($w_d$) to the solid line path ($w_s$), we analyze the two thermodynamic paths using the given $P$-$V$ graph.

1. Work Done Along the Solid Line Path ($w_s$)

From the graph, the solid curve represents an isothermal expansion process for an ideal gas where $P \cdot V = \text{constant}$:

• At point $\mathbf{a}$: $P_a = 4.0\text{ atm}$ and $V_a = 0.5\text{ L}$, so $P_a V_a = 2.0\text{ atm}\cdot\text{L}$.
• At intermediate points indicated by circles:
  • $P = 2.0\text{ atm}, V = 1.0\text{ L} \implies PV = 2.0\text{ atm}\cdot\text{L}$
  • $P = 1.0\text{ atm}, V = 2.0\text{ L} \implies PV = 2.0\text{ atm}\cdot\text{L}$
  • $P = 0.5\text{ atm}, V = 4.0\text{ L} \implies PV = 2.0\text{ atm}\cdot\text{L}$
• At point $\mathbf{b}$: $V_b = 5.5\text{ L}$.

The work done along an isothermal process is given by:
$w_s = \int_{V_a}^{V_b} P \, dV = P_a V_a \ln\left(\frac{V_b}{V_a}\right)$

Substituting the values:
$w_s = 2.0 \ln\left(\frac{5.5}{0.5}\right) = 2 \ln(11)$

Since $\ln(11) \approx 2.398$:
$w_s \approx 2 \times 2.398 = 4.796\text{ atm}\cdot\text{L}$

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2. Work Done Along the Dashed Line Path ($w_d$)

The dashed path consists of horizontal (isobaric expansion) and vertical (isochoric) segments:

1. First isobaric expansion at $P_1 = 4.0\text{ atm}$ from $V = 0.5\text{ L}$ to $V = 2.0\text{ L}$:
   $w_1 = P_1 \Delta V_1 = 4.0 \times (2.0 - 0.5) = 6.0\text{ atm}\cdot\text{L}$

2. First isochoric pressure drop at $V = 2.0\text{ L}$ from $P = 4.0\text{ atm}$ to $P = 1.0\text{ atm}$:
   $w_2 = 0$

3. Second isobaric expansion at $P_2 = 1.0\text{ atm}$ from $V = 2.0\text{ L}$ to $V = 3.0\text{ L}$:
   $w_3 = P_2 \Delta V_2 = 1.0 \times (3.0 - 2.0) = 1.0\text{ atm}\cdot\text{L}$

4. Second isochoric pressure drop at $V = 3.0\text{ L}$ from $P = 1.0\text{ atm}$ to $P_3 \approx \frac{2}{3}\text{ atm} \approx 0.67\text{ atm}$:
   $w_4 = 0$

5. Third isobaric expansion at $P_3 \approx \frac{2}{3}\text{ atm}$ from $V = 3.0\text{ L}$ to $V = 5.5\text{ L}$:
   $w_5 = P_3 \Delta V_3 = \frac{2}{3} \times (5.5 - 3.0) = \frac{2}{3} \times 2.5 = \frac{5}{3} \approx 1.67\text{ atm}\cdot\text{L}$

Summing the work done along each step:
$w_d = w_1 + w_2 + w_3 + w_4 + w_5 = 6.0 + 0 + 1.0 + 0 + 1.67 = 8.67\text{ atm}\cdot\text{L}$

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3. Ratio $w_d / w_s$

$\frac{w_d}{w_s} = \frac{8.67}{4.796} \approx 1.81$

The integer closest to $1.81$ is $2$.