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Heat and Thermodynamics question

2020 · 8 Jan · Shift 2 · Q43
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Heat and Thermodynamics question

2020 · 8 Jan · Shift 2 · Q43

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
Consider a mixture of n moles of helium gas and 2n moles of oxygen gas (molecules taken to be rigid) as an ideal gas. Its CP/CV value will be :
  1. A
    23/15
  2. B
    67/45
  3. C
    40/27
  4. D
    19/13
View written solutionFree

Correct answer: D

  1. Find degrees of freedom and molar heat capacities of each gas

For an ideal gas with molecules taken as rigid:

  • Helium is monatomic, so degrees of freedom f=3f=3f=3. CV(He)=32R,CP(He)=52RC_V(\text{He})=\frac{3}{2}R, \qquad C_P(\text{He})=\frac{5}{2}RCV​(He)=23​R,CP​(He)=25​R

  • Oxygen is diatomic and rigid, so degrees of freedom f=5f=5f=5. CV(O2)=52R,CP(O2)=72RC_V(\text{O}_2)=\frac{5}{2}R, \qquad C_P(\text{O}_2)=\frac{7}{2}RCV​(O2​)=25​R,CP​(O2​)=27​R

  1. Compute total CVC_VCV​ of the mixture

Given:

  • Helium = nnn moles
  • Oxygen = 2n2n2n moles

So total heat capacity at constant volume is: CV(mix)=n(32R)+2n(52R)C_V^{\text{(mix)}}=n\left(\frac{3}{2}R\right)+2n\left(\frac{5}{2}R\right)CV(mix)​=n(23​R)+2n(25​R)

=32nR+5nR=\frac{3}{2}nR+5nR=23​nR+5nR

=132nR=\frac{13}{2}nR=213​nR

  1. Compute total CPC_PCP​ of the mixture

CP(mix)=n(52R)+2n(72R)C_P^{\text{(mix)}}=n\left(\frac{5}{2}R\right)+2n\left(\frac{7}{2}R\right)CP(mix)​=n(25​R)+2n(27​R)

=52nR+7nR=\frac{5}{2}nR+7nR=25​nR+7nR

=192nR=\frac{19}{2}nR=219​nR

  1. Find the ratio γ=CPCV\gamma = \dfrac{C_P}{C_V}γ=CV​CP​​
=\frac{\frac{19}{2}nR}{\frac{13}{2}nR} =\frac{19}{13}$$ 5. **Match with the options** $$\boxed{\frac{19}{13}}$$ So the correct option is **D**.
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