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Creating an Energy Efficient Engine

A gas undergo...

Question

A gas undergoes $A$ to $B$ through three different processes $1, 2$ and $3$ as shown in the figure. The heat supplied to the gas is $Q_{1}, Q_{2}$ and $Q_{3}$ respectively, then

Q_{1} = Q_{2} = Q_{3}

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Q_{1} < Q_{2} < Q_{3}

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Q_{1} > Q_{2} > Q_{3}

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Q_{1} = Q_{3} > Q_{2}

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Solution

The correct option is C $Q_{1} > Q_{2} > Q_{3}$
As the initial and final thermodynamics state is same for all the three processes,
So, $Δ U$ will be equal for all three processes.

Thus, $Δ U_{1} = Δ U_{2} = Δ U_{3} = U_{B} - U_{A}$

So, let's assume $U_{B} - U_{A} = c$

From first law of thermodynamics,

$Δ Q = Δ U + Δ W$

$\Rightarrow Δ Q = c + Δ W . . . . . . . . . . . (1)$

In process $1$ ,
the process is in clockwise direction.

$∴ Δ W_{1} = + v e . . . . . . . . . (i)$

and similarly, in process $3$ ,
the process is in anticlockwise direction.

$∴ Δ W_{3} = - v e . . . . . . . . . . . (i i)$

In process $2, Δ V = 0$ ,

Thus, $Δ W_{2} = 0 . . . . . . . . . . . . . (i i i)$

So, from eqs. (i), (ii) and (iii)

$Δ W_{1} > Δ W_{2} > Δ W_{3}$

Now, from Eq. (1), we get

$Q_{1} > Q_{2} > Q_{3}$

Hence, option $(c)$ is the correct answer.

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