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An ideal gas expands from $10^{-3} m^3$ to $10^{-2} m^3$ at 300 K, against a constant pressure of $10^5 N m^{-2}$. The work done on it is

(a) -0.9 kJ
(b) -900 kJ
(c) 270 kJ
(d) -270 kJ
Can you answer this question?
 
 

1 Answer

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Answer: -0.9 kJ
$W = - P \Delta V$
$P = 10^5 Nm^{-2}$ = 1 atm
$\Delta V$ = 10 -1 = 9 Litres [$1m^3 = 10^3 L$]
W = $-\frac{1 \times 9 \times 101}{1000}$ [1 L-atom = 101 J]
W = 0.9 kJ
answered Feb 24, 2014 by mosymeow_1
 

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