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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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Two moles of a perfect gas undergoes following processes: \[\] 1. A reversible isobaric expansion from (1 atm, 20L) to (1 atm, 40L)\[\] 2. A reversible isochoric change of state from (1 atm, 40L) to (0.5 atm, 40L)\[\] 3. A reversible isothermal compression from (0.5 atm, 40L) to (1atm, 20L)\[\] Calculate the total work done involved in the above processes

$\begin {array} {1 1} (A)\;343\: J & \quad (B)\;-620.96 \: J \\ (C)\;-343\: J & \quad (D)\;620.96 \: J \end {array}$

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Total work done $W$ is given as $W = W_1 + W_2 + W_3$
Now, $W_1 = -P \Delta V = -1 \times (40-20) = -20\: L\: -atm$
$W_2 = 0$ because $ \Delta V = 0$
$W_3 = -2.303nRT \: log \large\frac{V_2}{V_1}$$ = -2.303 \times 2 \times 0.082 \times 121.95 \times log \bigg(\large\frac{20}{40} \bigg)$
$ = 13.86\: L\: -atm$
$W = W_1 + W_2 + W_3 = -20 + 0 + 13.86 = 6.13\: L\: -atm \times 101.3 = 620.96 \: J$
Ans : (D)
answered Mar 20, 2014 by thanvigandhi_1
 

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