Three moles of an ideal monoatomic gas undergoes a cyclic process as shown in the figure. The temperature of the gas in different states marked as 1, 2, 3 and 4 are $400 K, 700 K, 2500 K$ and $1100 K$ respectively. The work done by the gas during the process 1-2-3-4-1 is :
(Universal gas constant $R = 8.314 J / m o l / K$ )

1650 R

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550 R

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1100 R

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2200 R

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Solution

The correct option is A $1650 R$
From Ideal Gas Equation, $P V = n R T$

Work done in the process is given by the area enclosed in the plot.
Area of the plot is given by the area of the trapezium 1234.
$A_{1234} = A_{O B 3} - A_{O B 2} - A_{O A 4} + A_{O A 1}$
$= \frac{1}{2} P_{3} V_{3} - \frac{1}{2} P_{2} V_{2} - \frac{1}{2} P_{4} V_{4} + \frac{1}{2} P_{1} V_{1}$
$= \frac{1}{2} n R (T_{3} + T_{1} - T_{2} - T_{4})$
$= \frac{1}{2} \times 3 \times R \times (2500 + 400 - 700 - 1100)$
$= 1650 R$

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3

A B C D A

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

T_{D} = 1200 K

(R = 8.314 J / m o l e K)

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]

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

T_{1} = 400 K, T_{2} = 800 K, T_{3} = 2400 K a n d T_{4} = 1200 K

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