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Heat and Thermodynamics question

2020 · 6 Sep · Shift 1 · Q42
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Heat and Thermodynamics question

2020 · 6 Sep · Shift 1 · Q42

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom.The gas is maintained at a temperature of T. The total internal energy, U of a mole of this gas, and the value of γ(=CpCv)\gamma \left( { = {{{C_p}} \over {{C_v}}}} \right)γ(=Cv​Cp​​) are given, respectively by:
  1. A
    U = 52RT{5 \over 2}RT25​RT and γ=75\gamma = {7 \over 5}γ=57​
  2. B
    U = 5RT and γ=65\gamma = {6 \over 5}γ=56​
  3. C
    U = 5RT and γ=75\gamma = {7 \over 5}γ=57​
  4. D
    U = 52RT{5 \over 2}RT25​RT and γ=65\gamma = {6 \over 5}γ=56​
View written solutionFree

Correct answer: A

  1. Given degrees of freedom

For the ideal gas molecule:

  • Translational degrees of freedom =3= 3=3
  • Rotational degrees of freedom =2= 2=2

So total active degrees of freedom, f=3+2=5f = 3 + 2 = 5f=3+2=5

  1. Internal energy of 1 mole of an ideal gas

For an ideal gas, U=f2RTU = \frac{f}{2}RTU=2f​RT for one mole.

Substituting f=5f=5f=5, U=52RTU = \frac{5}{2}RTU=25​RT

  1. Molar specific heat at constant volume

For an ideal gas, Cv=f2R=52RC_v = \frac{f}{2}R = \frac{5}{2}RCv​=2f​R=25​R

  1. Molar specific heat at constant pressure

Using Cp=Cv+RC_p = C_v + RCp​=Cv​+R we get Cp=52R+R=72RC_p = \frac{5}{2}R + R = \frac{7}{2}RCp​=25​R+R=27​R

  1. Calculate γ\gammaγ

γ=CpCv=72R52R=75\gamma = \frac{C_p}{C_v} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5}γ=Cv​Cp​​=25​R27​R​=57​

  1. Match with the options

We found:

  • U=52RTU = \frac{5}{2}RTU=25​RT
  • γ=75\gamma = \frac{7}{5}γ=57​

This matches Option A.

  1. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So, the derived answer agrees with the stored correct answer.

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