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Heat and Thermodynamics question

2022 · 29 Jun · Shift 1 · Q52
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  5. /2022 · 29 Jun · Shift 1 · Q52

Heat and Thermodynamics question

2022 · 29 Jun · Shift 1 · Q52

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
In van der Waal equation [P+aV2]\left[ {P + {a \over {{V^2}}}} \right][P+V2a​][V −-− b] = RT; P is pressure, V is volume, R is universal gas constant and T is temperature. The ratio of constants ab{a \over b}ba​ is dimensionally equal to :
  1. A
    PV{P \over V}VP​
  2. B
    VP{V \over P}PV​
  3. C
    PV
  4. D
    PV3
View written solutionFree

Correct answer: C

  1. Write the van der Waals equation

    (P+aV2)(V−b)=RT\left(P+\frac{a}{V^2}\right)(V-b)=RT(P+V2a​)(V−b)=RT

  2. Find the dimensions of aaa

    In the bracket, P+aV2P+\frac{a}{V^2}P+V2a​ both terms must have the same dimensions.

    Therefore, aV2∼P\frac{a}{V^2} \sim PV2a​∼P

    So, a∼P V2a \sim P\,V^2a∼PV2

  3. Find the dimensions of bbb

    In the factor (V−b)(V-b)(V−b), both terms must have the same dimensions.

    Therefore, b∼Vb \sim Vb∼V

  4. Find the dimensions of ab\dfrac{a}{b}ba​

    ab∼PV2V=PV\frac{a}{b} \sim \frac{P V^2}{V} = PVba​∼VPV2​=PV

  5. Match with the given options

    • A: PV\dfrac{P}{V}VP​
    • B: VP\dfrac{V}{P}PV​
    • C: PVPVPV
    • D: PV3PV^3PV3

    Hence, the correct option is: PV\boxed{PV}PV​

  6. Compare with stored correct answer

    Stored correct answer: C

    Derived answer: C

    They match.

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