Verified Solution • 2025–26 Session moderate

QJEEM19Chaab9: JEE Main 2019 Chemistry - Class 11 JEEM19Chaab9 Solution: Consider the reversible isothermal expansion of an ideal gas in …

Consider the reversible isothermal expansion of an ideal gas in a closed system at two different temperatures T1 and T2 (T1 < T2). The correct graphical depiction of the dependence of work done (w) on the final volume (V) is :

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/b22071c3ed8cc7923efb0e0f03733b38319fad4d07cdaa55bef2459ccd728e3b.webp
  • B https://ik.imagekit.io/novatra/photos/ab49515a6b0ed0bfa78ff2d486fc6d484fd76da76748aa6f25e917044ef30d9c.webp
  • C https://ik.imagekit.io/novatra/photos/baeebd524966ef70a18f86b0e325f7ee80981c296d6397dc11194e17524b6a84.webp
  • D https://ik.imagekit.io/novatra/photos/8711bba29341f1402d031ea37999bfdc26602e05d8a1cb3eeb7d439aa098e434.webp

Step-by-Step Verified Solution

Work done in isothermal process,

W = −- nRT ln V2V1{{{V_2}} \over {{V_1}}}

∴\therefore     ∣W∣\left| W \right| = nRT ln V2V1{{{V_2}} \over {{V_1}}}

⇒\Rightarrow     ∣W∣\left| W \right| = nRT (lnV2 −- lnV1)

⇒\Rightarrow     ∣W∣\left| W \right| = nRT lnV2 −- nRT lnV1

given that VV 2 == VV

∴\therefore     ∣W∣\left| W \right| = nRT lnV −- nRT lnV1

By comparing with the straight line equation,

y = mx + C

We get slope is nRT and intercept −- nRT lnV1 in ∣W∣\left| W \right| and lnV graph.

As T2 > T1   So,

Slope nRT2 > nRT1

and intercept

−- nRT2 lnV1 < −- nRT1 lnV1

So, we can say

(1)   slope of T2 line is more then T1

(2)  intercept of T1 line is less negative than T2 line, and intercept of T1 can't be positive, it can be 0 or less than 0 as −- nRT1lnV1 always ≤\le 0

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