Question

At T(K) 10 L of an ideal gas was expanded to 12 L against a pressure of 2 atm irreversibly. What is the work done by the gas?

• A. -4 atm L
• B. Zero
• C. -240 atm L
• D. -4 J

Hint:

Irreversible expansion: $W = -P_{ext} \Delta V$; reversible uses integral $-nRT \ln(V_f/V_i)$. Sign negative for expansion (system does work). Check units.

Answer 1

The Correct Option is A

1. Work in expansion: for irreversible process against constant external pressure, $W = - P_{ext} \Delta V$.
2. Given $\Delta V = 12 - 10 = 2$ L, $P_{ext} = 2$ atm.
3. Thus, $W = - 2 \times 2 = -4$ atm L.
4. The temperature T is given but not needed for irreversible work calculation.
5. Note: Units atm L; if converted to J, 1 atm L = 101.325 J, but option in atm L.
6. Therefore, the correct option is (1) -4 atm L.