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Q.

We make helium gas of $2 g$ to go through the thermal cycle shown in the figure. The lowest temperature of the gas in the cycle is $- 17^{\circ} C$ the highest is $127^{\circ} C$ and the temperature is equal at points $A$ and $B$ . [ $R =$ Universal gas constant]

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a

the temperature at $A$ & $B$ is $328 K$

b

net work done by the gas during a cycle is $32 R$

c

the temperature of $A$ & $B$ is $320 K$

d

net work done by the gas during a cycle is $8 R$

answer is C, D.

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Detailed Solution

Question Image

Process $D \to A$ is isochoric, $\frac{P_{0}}{256} = \frac{P}{T_{1}}$ … (1)

$C \to B$ is isochoric, $\frac{P_{0}}{400} = \frac{P}{T_{1}}$ … (2)

from equation (1) $\times$ equation (2)
we get $T_{1} = 320 K$

$W_{A \to B} = η R ∆ T = \frac{2}{4} R [400 - 320] = 40 R W_{A \to B} = η R (T_{D} - T_{B}) = - 32 R W_{n e t} = 8 R$

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