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Heat and Thermodynamics question

2021 · 20 Jul · Shift 2 · Q58
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  5. /2021 · 20 Jul · Shift 2 · Q58

Heat and Thermodynamics question

2021 · 20 Jul · Shift 2 · Q58

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The correct relation between the degrees of freedom f and the ratio of specific heat γ\gammaγ is :
  1. A
    f=2γ−1f = {2 \over {\gamma - 1}}f=γ−12​
  2. B
    f=2γ+1f = {2 \over {\gamma + 1}}f=γ+12​
  3. C
    f=γ+12f = {{\gamma + 1} \over 2}f=2γ+1​
  4. D
    f=1γ+1f = {1 \over {\gamma + 1}}f=γ+11​
View written solutionFree

Correct answer: A

  1. For an ideal gas with degrees of freedom fff:

CV=f2RC_V = \frac{f}{2}RCV​=2f​R

and

CP=CV+R=f2R+R=f+22RC_P = C_V + R = \frac{f}{2}R + R = \frac{f+2}{2}RCP​=CV​+R=2f​R+R=2f+2​R

  1. The ratio of specific heats is

γ=CPCV=f+22Rf2R=f+2f\gamma = \frac{C_P}{C_V} = \frac{\frac{f+2}{2}R}{\frac{f}{2}R} = \frac{f+2}{f}γ=CV​CP​​=2f​R2f+2​R​=ff+2​

  1. Simplify:

γ=1+2f\gamma = 1 + \frac{2}{f}γ=1+f2​

So,

γ−1=2f\gamma - 1 = \frac{2}{f}γ−1=f2​

  1. Rearranging for fff:

f=2γ−1f = \frac{2}{\gamma - 1}f=γ−12​

  1. Now compare with the options:
  • A: f=2γ−1f = \dfrac{2}{\gamma - 1}f=γ−12​ ✅
  • B: f=2γ+1f = \dfrac{2}{\gamma + 1}f=γ+12​ ❌
  • C: f=γ+12f = \dfrac{\gamma + 1}{2}f=2γ+1​ ❌
  • D: f=1γ+1f = \dfrac{1}{\gamma + 1}f=γ+11​ ❌

Hence, the correct option is A.

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