One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is ws and that along the dotted line path is wd, then the integer closet to the ratio wd / ws is

Engineering

Chemistry

Gas Laws and Graphical Representation

First Law and Various Process

Ideal Gas Equation

Question

One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is w_s and that along the dotted line path is w_d, then the integer closet to the ratio w_d / w_s is

For solid line path show approxy isothermal process

∴ work done |WS| = 2.303 (PV) log $\frac{5.5}{.5}$

=2.303 × 4 × .5 × log 11

$≃$ 4.79

for dashed line path work done

wd = 4 × |2 – .5| + 1 × |3 – 2| + .5 × |5.5 –3|

= 6 + 1 + 1.25 = 8.25

$∴ \frac{w_{d}}{w_{s}} = \frac{8 .25}{4 .8} = 1 .71 ≃ 2$

This problem involves calculating work done by an ideal gas during expansion along two different paths on a P-V diagram and finding their ratio. Work done by a gas is given by the area under the P-V curve.

Step 1: Understanding Work Calculation

For any thermodynamic process, work done by the gas (W) is given by:

W = \int P d V

On a P-V diagram, this represents the area under the curve from initial to final volume.

Step 2: Analyzing the Given Paths

The graph shows two paths from point a (2 L, 4 atm) to point b (6 L, 2 atm):

Solid path: A curved path (likely representing an isothermal process for an ideal gas, though not explicitly stated, but area calculation is key).
Dashed path: A two-step path: first an isobaric expansion (constant pressure) from (2L,4atm) to (6L,4atm), followed by an isochoric pressure reduction (constant volume) from (6L,4atm) to (6L,2atm).

Step 3: Calculating Work Along Dashed Path (w_d)

The dashed path consists of:

Isobaric expansion at P=4 atm from V=2L to V=6L: $W_{1} = P Δ V = 4 \times (6 - 2) = 16 L.atm$
Isochoric process: Volume constant, so work done is zero: $W_{2} = 0$

Total work for dashed path: $w_{d} = W_{1} + W_{2} = 16 + 0 = 16 L.atm$

Step 4: Calculating Work Along Solid Path (w_s)

The solid path is a smooth curve from a to b. For an ideal gas, if this is an isothermal process (constant temperature), work done is given by:

w_{s} = n R T \ln (\frac{V_{f}}{V_{i}})

Since n=1 mole, and using PV=nRT, we can find RT. At point a: P=4 atm, V=2 L, so: $R T = \frac{P V}{n} = \frac{4 \times 2}{1} = 8 L.atm$

Thus, $w_{s} = 1 \times 8 \times \ln (\frac{6}{2}) = 8 \times \ln (3)$

Since ln(3) ≈ 1.0986, $w_{s} \approx 8 \times 1.0986 \approx 8.789 L.atm$

Step 5: Finding the Ratio w_d / w_s

\frac{w_{d}}{w_{s}} = \frac{16}{8.789} \approx 1.820

The integer closest to 1.820 is 2.

Final Answer

The integer closest to the ratio w_d / w_s is 2.

Detailed Solution

Step 1: Understanding Work Calculation

Step 2: Analyzing the Given Paths

Step 3: Calculating Work Along Dashed Path (w_d)

Step 4: Calculating Work Along Solid Path (w_s)

Step 5: Finding the Ratio w_d / w_s

Final Answer

Related Topics & Formulae

Detailed Solution

Step 1: Understanding Work Calculation

Step 2: Analyzing the Given Paths

Step 3: Calculating Work Along Dashed Path (wd)

Step 4: Calculating Work Along Solid Path (ws)

Step 5: Finding the Ratio wd / ws

Final Answer

Related Topics & Formulae

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Step 3: Calculating Work Along Dashed Path (w_d)

Step 4: Calculating Work Along Solid Path (w_s)

Step 5: Finding the Ratio w_d / w_s