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I am told that the area under the PV curve gives the magnitude of work done by a gas.

Let $V_i$ be the initial volume, $V_f$ be the final volume, and the pressure as a function of volume be given by $P=f(V)$. To my understanding, the work done is given by the area bounded by $x=V_i$, $x=V_f$, $P=F(v)$ and $y=0$.

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Proceeding in the above stated manner, the work done comes out to be the sum of the areas of green and yellow region, but the answer stated neglects the yellow region.

Origin of Question : JEE MAIN

Here's the answer according to the official answer key released by the testing agency: enter image description here

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The error comes from reading the graph: the crossing of the axes corresponds to the point (2,2) and not (0,0)

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  • $\begingroup$ If that is correct, then we must be dealing with a thermodynamic cycle here, with the lower boundary of the cycle p = 2. Otherwise, where we placed the lower boundary on the graph would not matter. $\endgroup$ Feb 18 at 3:36
  • $\begingroup$ @ChetMiller but the question states that the inital state of gas is A and final state is C, which does not form a thermodynamic cycle. $\endgroup$ Feb 18 at 6:06
  • $\begingroup$ I still think that implicitly, it is necessary to add a point D (2,2) and that the gas describes the cycle ABCDA otherwise we cannot effectively neglect the yellow region in the added image $\endgroup$
    – Nicolas
    Feb 18 at 10:07
  • $\begingroup$ I agree with Nicolases'. And, if the yellow area is included, then the work is 64. $\endgroup$ Feb 18 at 11:34
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    $\begingroup$ @SuchetSaxena even the biggest can make mistakes! Nothing justifies the given value of $ \pu{48J} $ other than that of a cycle such as the one shown in the figure! $\endgroup$
    – Nicolas
    Feb 18 at 12:15

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