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Heat and Thermodynamics question

2023 · 31 Jan · Shift 2 · Q55
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  5. /2023 · 31 Jan · Shift 2 · Q55

Heat and Thermodynamics question

2023 · 31 Jan · Shift 2 · Q55

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A hypothetical gas expands adiabatically such that its volume changes from 08 litres to 27 litres. If the ratio of final pressure of the gas to initial pressure of the gas is 1681\frac{16}{81}8116​. Then the ratio of CpCv\frac{\mathrm{Cp}}{\mathrm{Cv}}CvCp​ will be.
  1. A
    31\frac{3}{1}13​
  2. B
    43\frac{4}{3}34​
  3. C
    12\frac{1}{2}21​
  4. D
    32\frac{3}{2}23​
View written solutionFree

Correct answer: B

  1. For an adiabatic process of an ideal gas,

PVγ=constantPV^\gamma = \text{constant}PVγ=constant

where

γ=CpCv\gamma = \frac{C_p}{C_v}γ=Cv​Cp​​

  1. Therefore, between initial and final states,

PiViγ=PfVfγP_i V_i^\gamma = P_f V_f^\gammaPi​Viγ​=Pf​Vfγ​

So,

PfPi=(ViVf)γ\frac{P_f}{P_i} = \left(\frac{V_i}{V_f}\right)^\gammaPi​Pf​​=(Vf​Vi​​)γ

  1. Substitute the given values:

Vi=8 L,Vf=27 L,PfPi=1681V_i = 8\text{ L}, \qquad V_f = 27\text{ L}, \qquad \frac{P_f}{P_i} = \frac{16}{81}Vi​=8 L,Vf​=27 L,Pi​Pf​​=8116​

Thus,

1681=(827)γ\frac{16}{81} = \left(\frac{8}{27}\right)^\gamma8116​=(278​)γ

  1. Rewrite both fractions in powers:

1681=2434=(23)4\frac{16}{81} = \frac{2^4}{3^4} = \left(\frac{2}{3}\right)^48116​=3424​=(32​)4

and

827=2333=(23)3\frac{8}{27} = \frac{2^3}{3^3} = \left(\frac{2}{3}\right)^3278​=3323​=(32​)3

So the equation becomes

(23)4=[(23)3]γ=(23)3γ\left(\frac{2}{3}\right)^4 = \left[\left(\frac{2}{3}\right)^3\right]^\gamma = \left(\frac{2}{3}\right)^{3\gamma}(32​)4=[(32​)3]γ=(32​)3γ

  1. Equate the powers:

3γ=43\gamma = 43γ=4

Hence,

γ=43\gamma = \frac{4}{3}γ=34​

  1. Since

γ=CpCv\gamma = \frac{C_p}{C_v}γ=Cv​Cp​​

the required ratio is

43\boxed{\frac{4}{3}}34​​

  1. Checking options:
  • A: 333 ❌
  • B: 43\frac{4}{3}34​ ✅
  • C: 12\frac{1}{2}21​ ❌
  • D: 32\frac{3}{2}23​ ❌

So the correct option is B.

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