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Question

Form first law of the thermodynamics $\mathrm{dQ}=\mathrm{dU}+\mathrm{d} \mathrm{W}$

In our case, $\mathrm{Q}=\mathrm{U}+\mathrm{W}$

Here, $\mat

Vedclass · Accepted Answer

$W_A>W_B> W_C$

Vedclass · Answer

Form first law of the thermodynamics $\mathrm{dQ}=\mathrm{dU}+\mathrm{d} \mathrm{W}$

In our case, $\mathrm{Q}=\mathrm{U}+\mathrm{W}$

Here, $\mathrm{U}(\mathrm{Q}-\mathrm{W})$ depends only on initial and final states, so $U$ will be same for all curves.

Wrepresents the area surrounded by the curve by work done.

so, $W_{A}>W_{B}>W_{C}$

 or       $\mathrm{Q}_{\mathrm{A}}>\mathrm{Q}_{\mathrm{B}}>\mathrm{Q}_{\mathrm{C}}$

Here, choice

$(2)$ is correct.

Given diagram shows an ideal gas taken from state $1$ to $2$ through optional paths, $A,B,C.$ Let $Q,W$ and $U$ represent the heat supplied to, the work done by gas and the internal energy of the gas, respectively. Then which of the following conditions is true?

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