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An ideal gas expands in volume from 1×10−31 \times 10^{-3} m3^{3} to 1×10−21 \times 10^{-2} m3^{3} at 300 K against a constant pressure of 1×1051 \times 10^{5} N m−2^{-2}. The work done is [AIEEE 2004]:

A

270 kJ

B

–900 kJ

C

–900 J

D

900 kJ

Step-by-Step Solution

To calculate the work done during the expansion of an ideal gas against a constant external pressure, we use the formula w=−pexΔVw = -p_{ex} \Delta V .

  1. Identify Given Values: Initial volume (ViV_{i}) = 1×10−31 \times 10^{-3} m3^{3}. Final volume (VfV_{f}) = 1×10−21 \times 10^{-2} m3^{3}.
  • Constant external pressure (pexp_{ex}) = 1×1051 \times 10^{5} N m−2^{-2} (which is equal to 1×1051 \times 10^{5} Pa) .

  1. Calculate the change in volume (ΔV\Delta V): ΔV=Vf−Vi=(1×10−2)−(1×10−3)\Delta V = V_{f} - V_{i} = (1 \times 10^{-2}) - (1 \times 10^{-3}) m3^{3}. ΔV=(10×10−3)−(1×10−3)=9×10−3\Delta V = (10 \times 10^{-3}) - (1 \times 10^{-3}) = 9 \times 10^{-3} m3^{3}.

  2. Calculate Work (ww): w=−(1×105 N m−2)×(9×10−3 m3)w = -(1 \times 10^{5} \text{ N m}^{-2}) \times (9 \times 10^{-3} \text{ m}^{3}). w=−9×102 N m=−900w = -9 \times 10^{2} \text{ N m} = -900 J .

According to IUPAC sign conventions, the negative sign indicates that work is done by the system during expansion . Therefore, the work done is –900 J, matching Option C.

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