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Heat and Thermodynamics question

2022 · 25 Jun · Shift 2 · Q51
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  5. /2022 · 25 Jun · Shift 2 · Q51

Heat and Thermodynamics question

2022 · 25 Jun · Shift 2 · Q51

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The ratio of specific heats (CPCV)\left( {{{{C_P}} \over {{C_V}}}} \right)(CV​CP​​) in terms of degree of freedom (f) is given by :
  1. A
    (1+f3)\left( {1 + {f \over 3}} \right)(1+3f​)
  2. B
    (1+2f)\left( {1 + {2 \over f}} \right)(1+f2​)
  3. C
    (1+f2)\left( {1 + {f \over 2}} \right)(1+2f​)
  4. D
    (1+1f)\left( {1 + {1 \over f}} \right)(1+f1​)
View written solutionFree

Correct answer: B

  1. For an ideal gas with degree of freedom fff, the molar specific heat at constant volume is CV=f2RC_V = \frac{f}{2}RCV​=2f​R because each degree of freedom contributes 12R\frac{1}{2}R21​R per mole.

  2. For an ideal gas, CP=CV+RC_P = C_V + RCP​=CV​+R So, CP=f2R+R=(f2+1)RC_P = \frac{f}{2}R + R = \left(\frac{f}{2}+1\right)RCP​=2f​R+R=(2f​+1)R

  3. Now the ratio of specific heats is γ=CPCV\gamma = \frac{C_P}{C_V}γ=CV​CP​​ Substituting the values: γ=(f2+1)Rf2R\gamma = \frac{\left(\frac{f}{2}+1\right)R}{\frac{f}{2}R}γ=2f​R(2f​+1)R​

  4. Cancel RRR: γ=f2+1f2\gamma = \frac{\frac{f}{2}+1}{\frac{f}{2}}γ=2f​2f​+1​

  5. Simplify: γ=1+2f\gamma = 1 + \frac{2}{f}γ=1+f2​

  6. Comparing with the options:

  • A: 1+f31+\frac{f}{3}1+3f​
  • B: 1+2f1+\frac{2}{f}1+f2​
  • C: 1+f21+\frac{f}{2}1+2f​
  • D: 1+1f1+\frac{1}{f}1+f1​

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

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