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3 moles of an ideal monoatomic gas performs ABCDA cyclic process as shown in figure below. The gas temperatures are $T_{A} = 400 K, T_{B} = 800 K, T_{C} = 2400 K and T_{D} = 1200 K$ . The work done by the gas is (approximately) ( $R = 8.314 J / m o l e K$ )

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100 kJ

20 kJ

40 kJ

10 kJ

answer is B.

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Detailed Solution

For process $A B = W = 0$

For process $BC = W = nRΔT = 3 \times R [2400 - 800]$

$= 3 R \times 1600 = 4800 R$

For process CD = W=0

For process $D A = W = 3 R [400 - 1200] = - 2400 R$

Total work = $2400 R = 2400 \times \frac{25}{3} = 800 \times 25 = 20 kJ$

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