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1st Law of Thermodynamics

Four identica...

Question

Four identical samples of an ideal gas are taken along the paths $A, B, C$ and $D$ from state $1$ to state $2$ as shown in the $P V$ indicator diagram. Let $Q, W$ and $Δ U$ represent the heat supplied, work done and change in internal energy of the gas respectively. Choose the correct option among the following :

A

Q_{A} - Q_{D} = W_{A} - W_{D}

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B

Q_{B} - W_{B} > Q_{C} - W_{C}

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C

W_{A} < W_{B} < W_{C} < W_{D}

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D

Q_{A} < Q_{B} < Q_{C} < Q_{D}

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Solution

The correct option is A $Q_{A} - Q_{D} = W_{A} - W_{D}$
We know that area under $P - V$ curve gives the work done during the process.
Work done by the gas in all four processes is positive and is in the order
$W_{A} > W_{B} > W_{C} > W_{D} . . . . . (1)$
Hence, option (c) is wrong.
The change in internal energy $Δ U$ is the same for all processes as initial and final states for all processes are the same.
Using first law of thermodynamics for each process $A, B, C$ and $D$ , we get
$Q_{A} = Δ U + W_{A} . . . . . . (2)$
$Q_{B} = Δ U + W_{B} . . . . . . (3)$
$Q_{C} = Δ U + W_{C} . . . . . . (4)$
$Q_{D} = Δ U + W_{D} . . . . . . (5)$
Hence, from $(1)$ we can say that, $Q_{A} > Q_{B} > Q_{C} > Q_{D}$
So, option (d) is also wrong.

From equation $(3)$ and $(4)$ ,
$Q_{B} - Q_{C} = W_{B} - W_{C}$
$\Rightarrow Q_{B} - W_{B} = Q_{C} - W_{C}$
i.e option (b) is wrong.

From equation $(2)$ and $(5)$ ,
$Q_{A} - Q_{D} = W_{A} - W_{D}$
Thus, option (a) is the correct answer.

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