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Heat and Thermodynamics question

2003 · Shift 0 · Q155
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Heat and Thermodynamics question

2003 · Shift 0 · Q155

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio Cp/CV{C_p}/{C_V}Cp​/CV​ for the gas is
  1. A
    43{4 \over 3}34​
  2. B
    222
  3. C
    53{5 \over 3}35​
  4. D
    32{3 \over 2}23​
View written solutionFree

Correct answer: D

  1. For an ideal gas in an adiabatic process,

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

where

γ=CpCV.\gamma=\frac{C_p}{C_V}.γ=CV​Cp​​.

  1. Using the ideal gas law,

PV=nRT  ⟹  V=nRTP.PV=nRT \implies V=\frac{nRT}{P}.PV=nRT⟹V=PnRT​.

Substitute this into the adiabatic relation:

P(nRTP)γ=constant.P\left(\frac{nRT}{P}\right)^{\gamma}=\text{constant}.P(PnRT​)γ=constant.

  1. Simplify:

P⋅(nR)γTγP−γ=constantP\cdot (nR)^{\gamma} T^{\gamma} P^{-\gamma}=\text{constant}P⋅(nR)γTγP−γ=constant

(nR)γTγP1−γ=constant. (nR)^{\gamma} T^{\gamma} P^{1-\gamma}=\text{constant}.(nR)γTγP1−γ=constant.

Ignoring constants,

TγP1−γ=constant.T^{\gamma}P^{1-\gamma}=\text{constant}.TγP1−γ=constant.

So,

Pγ−1∝Tγ.P^{\gamma-1}\propto T^{\gamma}.Pγ−1∝Tγ.

Hence,

P∝Tγγ−1.P\propto T^{\frac{\gamma}{\gamma-1}}.P∝Tγ−1γ​.

  1. Given in the question:

P∝T3.P\propto T^3.P∝T3.

Therefore,

γγ−1=3.\frac{\gamma}{\gamma-1}=3.γ−1γ​=3.

  1. Solve for γ\gammaγ:

γ=3(γ−1)\gamma=3(\gamma-1)γ=3(γ−1)

γ=3γ−3\gamma=3\gamma-3γ=3γ−3

2γ=32\gamma=32γ=3

γ=32.\gamma=\frac{3}{2}.γ=23​.

  1. Since

γ=CpCV,\gamma=\frac{C_p}{C_V},γ=CV​Cp​​,

we get

CpCV=32.\frac{C_p}{C_V}=\frac{3}{2}.CV​Cp​​=23​.

  1. Option check:
  • A: 43\frac{4}{3}34​ — incorrect
  • B: 222 — incorrect
  • C: 53\frac{5}{3}35​ — incorrect
  • D: 32\frac{3}{2}23​ — correct

Therefore, the correct answer is Option D.

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