Question

Calculate the work done in joule if 2 moles of an ideal gas expand isothermally from \(15.5 \text{ dm}^3\) to \(20 \text{ dm}^3\) at constant pressure 1 atm .

• A. -456 J
• B. -228 J
• C. -684 J
• D. -912 J

Hint:

For constant-pressure expansion: \[ w=-P\Delta V \] and always remember: \[ 1\ \text{L atm} = 101.3\ \text{J} \]

Answer 1

The Correct Option is A

Concept:
At constant external pressure, work done is: \[ w = -P\Delta V \] Here negative sign shows work done by the system during expansion.

Step 1: Calculate change in volume.
\[ \Delta V = V_2 - V_1 = 20 - 15.5 = 4.5\ \text{dm}^3 \]

Step 2: Use pressure-volume work formula.
\[ w = -P\Delta V = -(1\ \text{atm})(4.5\ \text{dm}^3) \] \[ w = -4.5\ \text{L atm} \]

Step 3: Convert \(\text{L atm}\) into joule.
\[ 1\ \text{L atm} = 101.3\ \text{J} \] So, \[ w = -4.5 \times 101.3 \] \[ w \approx -456\ \text{J} \] Hence, the correct answer is:
\[ \boxed{(A)\ -456\ \text{J}} \]