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Heat and Thermodynamics question

2025 · 2 Apr · Shift 1 · Q65
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Heat and Thermodynamics question

2025 · 2 Apr · Shift 1 · Q65

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The equation for real gas is given by (P+aV2)(V−b)=RT\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}(P+V2a​)(V−b)=RT, where P,V,T\mathrm{P}, \mathrm{V}, \mathrm{T}P,V,T and R are the pressure, volume, temperature and gas constant, respectively. The dimension of ab−2\mathrm{ab}^{-2}ab−2 is equivalent to that of :
  1. A
    Compressibility
  2. B
    Planck's constant
  3. C
    Energy density
  4. D
    Strain
View written solutionFree

Correct answer: C

  1. Given van der Waals type equation

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

We need the dimensions of ab−2ab^{-2}ab−2.

  1. Find dimension of aaa

In

P+aV2P+\frac{a}{V^2}P+V2a​

the two terms must have the same dimensions. Hence,

[aV2]=[P]\left[\frac{a}{V^2}\right]=[P][V2a​]=[P]

So,

[a]=[P][V]2[a]=[P][V]^2[a]=[P][V]2

Now,

[P]=ML−1T−2[P]=ML^{-1}T^{-2}[P]=ML−1T−2

and

[V]=L3[V]=L^3[V]=L3

Therefore,

[a]=(ML−1T−2)(L3)2=(ML−1T−2)(L6)=ML5T−2[a]=(ML^{-1}T^{-2})(L^3)^2=(ML^{-1}T^{-2})(L^6)=ML^5T^{-2}[a]=(ML−1T−2)(L3)2=(ML−1T−2)(L6)=ML5T−2

  1. Find dimension of bbb

From (V−b)(V-b)(V−b), both terms must have same dimensions, so

[b]=[V]=L3[b]=[V]=L^3[b]=[V]=L3

Thus,

[b−2]=L−6[b^{-2}]=L^{-6}[b−2]=L−6

  1. Find dimension of ab−2ab^{-2}ab−2

[ab−2]=[a][b−2][ab^{-2}]=[a][b^{-2}][ab−2]=[a][b−2]

=(ML5T−2)(L−6)=ML−1T−2=\left(ML^5T^{-2}\right)\left(L^{-6}\right)=ML^{-1}T^{-2}=(ML5T−2)(L−6)=ML−1T−2

  1. Identify the physical quantity

The dimension obtained is

ML−1T−2ML^{-1}T^{-2}ML−1T−2

Now compare with the options:

  • Compressibility: reciprocal of pressure [compressibility]=[P]−1=M−1LT2[\text{compressibility}] = [P]^{-1} = M^{-1}LT^2[compressibility]=[P]−1=M−1LT2 Not matching.

  • Planck's constant: [h]=ML2T−1[h]=ML^2T^{-1}[h]=ML2T−1 Not matching.

  • Energy density = energy/volume [energy]=ML2T−2[\text{energy}] = ML^2T^{-2}[energy]=ML2T−2 [energy density]=ML2T−2L3=ML−1T−2[\text{energy density}] = \frac{ML^2T^{-2}}{L^3}=ML^{-1}T^{-2}[energy density]=L3ML2T−2​=ML−1T−2 Matches.

  • Strain: dimensionless Not matching.

  1. Final answer

[ab−2]=ML−1T−2[ab^{-2}] = ML^{-1}T^{-2}[ab−2]=ML−1T−2 which is the dimension of energy density.

Therefore, the correct option is C.

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