$3$ moles of an ideal monoatomic gas performs $A B C D A$ cyclic process as shown in figure below. The gas temperature 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) $(R = 8.314 J / m o l e K)$ :

10 k J

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20 k J

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40 k J

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100 k J

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Solution

The correct option is A $10 k J$
$\begin{matrix} F o r p r o c e s s A B : w = 0 F o r p r o c e s s B C : w = n R Δ T = 3 \times R \times [2400 - 800] = 3 \times R \times 1600 = 4800 R F o r p r o c e s s C D : w = 0 F o r p r o c e s s D A : w = 3 R [400 - 1200] = - 2400 R ∴ T o t a l w o r k = 2400 R = 2400 \times \frac{25}{3} J = 800 \times 25 J = 10 K J \end{matrix}$
Hence,
option $(A)$ is correct answer.

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Similar questions

T_{A} = 400 K, T_{B} = 800 K; T_{C} = 2400 K; T_{D} = 1200 K

T_{A} = 400 K, T_{B} = 800 K, T_{C} = 2400 K, T_{D} = 1200 K .

3

A, B, C, D

T_{A} = 400 K,

T_{B} = 800 K, T_{C} = 2400 K, T_{D} = 1200 K .

T_{A} = 600 K, T_{B} = 800 K, T_{C} = 2200 K

T_{D} = 1200 K

A B C D A

T_{1} = 400 K, T_{2} = 800 K, T_{3} = 2400 K,

T_{4} = 1200 K

[R = \frac{25}{3} J / m o l - k]

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