Question

The work done during the expansion of a gas from a volume of $4\, dm^3$ to $6 \,dm^3$ against a constant external pressure of $3$ atm is :-

• A. -608 J
• B. #ERROR!
• C. -304 J
• D. -6 J

Answer 1

The Correct Option is A

To solve the problem of calculating the work done during the expansion of a gas, we use the formula for work done during an isobaric (constant pressure) process. The formula is:

\(W = - P_{\text{ext}} \times \Delta V\)

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, \(P_{\text{ext}} = 3 \, \text{atm}\)

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

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

Convert this volume change from \(\text{dm}^3\) to \(\text{L}\) since they are equivalent:

\(\Delta V = 2 \, \text{L}\)

The work done is typically expressed in Joules (J). First, convert the pressure from atm to Pa (Pascals). Remember that \(1 \, \text{atm} = 101325 \, \text{Pa}\):

\(P_{\text{ext}} = 3 \, \text{atm} = 3 \times 101325 \, \text{Pa} = 303975 \, \text{Pa}\)

Finally, calculate the work done using the formula:

\(W = - 303975 \, \text{Pa} \times 2 \times 10^{-3} \, \text{m}^3\)

Remember to convert the volume from liters to cubic meters (\(1 \, \text{L} = 10^{-3} \, \text{m}^3\)):

\(W = - 303975 \, \text{Pa} \times 0.002 \, \text{m}^3\)

Calculating this gives:

\(W = - 607.95 \, \text{J}\)

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

\(W = -608 \, \text{J}\)

Therefore, the correct answer is -608 J.