Question

The work done during the expansion of a gas from a volume of 4 dm³ to 6 dm³ against a constant external pressure of 3 atm is

• A. –608 J
• B. + 304 J
• C. –304 J
• D. –6 J

Answer 1

The Correct Option is A

To determine the work done during the expansion of a gas, we can use the formula for work done during gas expansion against a constant external pressure:

W = - P_{\text{ext}} \cdot \Delta V

where:

• W is the work done,
• P_{\text{ext}} is the external pressure,
• \Delta V is the change in volume.

Given:

• Initial volume V_1 = 4 \, \text{dm}^3
• Final volume V_2 = 6 \, \text{dm}^3
• External pressure P_{\text{ext}} = 3 \, \text{atm}

The change in volume, \Delta V is given by:

\Delta V = V_2 - V_1 = 6 - 4 = 2 \, \text{dm}^3

Since 1 dm³ = 1 litre and 1 litre = 1 x 10^-3 m³, we convert \Delta V to m³:

\Delta V = 2 \, \text{dm}^3 = 2 \times 10^{-3} \, \text{m}^3

The external pressure (P_{\text{ext}}) in SI units is given by:

1 \, \text{atm} = 101325 \, \text{Pa}

Therefore, P_{\text{ext}} = 3 \times 101325 = 303975 \, \text{Pa}

Substitute these values into the work done formula:

W = - 303975 \times 2 \times 10^{-3}

W = -607.95 \, \text{J}

Rounding to the nearest whole number, the work done is approximately:

W = -608 \, \text{J}

Therefore, the correct answer is –608 J.