An ideal gas is taken from state 1 to state 2 | vedclass

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(D) The work done $W$ is equal to the area under the $P-V$ curve.From the graph,the area under path $A$ is the largest,followed by $B$,$C$,and $D$ is

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Both $(B)$ and $(C)$

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(D) The work done $W$ is equal to the area under the $P-V$ curve.From the graph,the area under path $A$ is the largest,followed by $B$,$C$,and $D$ is the smallest.Thus,$W_A > W_B > W_C > W_D$ $...(1)$From the first law of thermodynamics,$Q = \Delta U + W$,which implies $Q - W = \Delta U$ $...(2)$Since $\Delta U$ is a state function,the change in internal energy is the same for all paths between state $1$ and state $2$.Therefore,$Q_A - W_A = Q_B - W_B = Q_C - W_C = Q_D - W_D = \Delta U$.Comparing this with the options:Option $(B)$ states $Q_A - W_A = Q_D - W_D$,which is correct as both equal $\Delta U$.Option $(C)$ states $Q_A > Q_B > Q_C > Q_D$. Since $Q = \Delta U + W$ and $\Delta U$ is constant,$Q$ follows the same order as $W$. Thus,$Q_A > Q_B > Q_C > Q_D$ is also correct.Since both $(B)$ and $(C)$ are correct,the correct option is $(D)$.

List-$I$	List-$II$
$A$. Pressure varies inversely with volume of an ideal gas.	$I$. Adiabatic process
$B$. Heat absorbed goes partly to increase internal energy and partly to do work.	$II$. Isochoric process
$C$. Heat is neither absorbed nor released by a system.	$III$. Isothermal process
$D$. No work is done on or by a gas.	$IV$. Isobaric process

An ideal gas is taken from state $1$ to state $2$ through optional paths $A, B, C$ and $D$ as shown in the $P-V$ diagram. Let $Q, W$ and $\Delta U$ represent the heat supplied,work done,and change in internal energy of the gas,respectively. Then:

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