Question

At 27^C, 1.6 g of O$_2$ gas at 5 atm expands isothermally against a constant external pressure of 1 atm. The work done (in J) is (1 L-atm = 100 J):

• A. -49.2
• B. -98.4
• C. +98.4
• D. +49.2

Hint:

Always use external pressure for irreversible work calculation: \(W = -P_{\text{ext}}\Delta V\).

Answer 1

The Correct Option is B

Step 1: Choose the right work formula.
The gas expands against a fixed outside pressure, so this is irreversible work. For such a case \[ W = -P_{ext}(V_2 - V_1) \] The minus sign shows the gas loses energy as it pushes outward.

Step 2: Find the moles of oxygen.
Oxygen has molar mass $32$ g/mol, so \[ n = \frac{1.6}{32} = 0.05 \text{ mol} \]

Step 3: Find the starting volume.
Using $V = \dfrac{nRT}{P}$ with $R = 0.082$, $T = 300$ K, and the initial pressure $5$ atm: \[ V_1 = \frac{0.05 \times 0.082 \times 300}{5} = 0.246 \text{ L} \]

Step 4: Find the final volume.
The gas expands until its pressure matches the outside pressure of $1$ atm: \[ V_2 = \frac{0.05 \times 0.082 \times 300}{1} = 1.23 \text{ L} \]

Step 5: Plug into the work formula.
\[ W = -1 \times (1.23 - 0.246) = -0.984 \text{ L atm} \]

Step 6: Convert to joules.
Using $1$ L atm $= 100$ J: \[ W = -0.984 \times 100 = -98.4 \text{ J} \] The negative sign confirms work is done by the gas on the surroundings.
\[ \boxed{-98.4 \text{ J}} \]