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An ideal gas is allowed to expand from 1 L to 10 L against a constant external pressure of 1 bar. The work done in kJ is:
(1) $-9.0$
(2) $+10.0$
(3) $-0.9$
(4) $-2.0$
$w = -P\Delta V = -(1 \text{ bar}) \times (9 \text{ L}) = -(10^5 \text{ Pa}) \times (9 \times 10^{-3}) \text{ m}^3 = -9 \times 10^2$ N.m $= -900$ J $= -0.9$ kJ.
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