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Heat and Thermodynamics question

2024 · 30 Jan · Shift 2 · Q73
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Heat and Thermodynamics question

2024 · 30 Jan · Shift 2 · Q73

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
If three moles of monoatomic gas (γ=53)\left(\gamma=\frac{5}{3}\right)(γ=35​) is mixed with two moles of a diatomic gas (γ=75)\left(\gamma=\frac{7}{5}\right)(γ=57​), the value of adiabatic exponent γ\gammaγ for the mixture is
  1. A
    1.35
  2. B
    1.52
  3. C
    1.40
  4. D
    1.75
View written solutionFree

Correct answer: B

  1. Use heat capacities of each gas

For an ideal gas,

γ=CpCv,Cp−Cv=R\gamma = \frac{C_p}{C_v}, \qquad C_p - C_v = Rγ=Cv​Cp​​,Cp​−Cv​=R

So,

Cv=Rγ−1,Cp=γRγ−1C_v = \frac{R}{\gamma - 1}, \qquad C_p = \frac{\gamma R}{\gamma - 1}Cv​=γ−1R​,Cp​=γ−1γR​
  1. Monoatomic gas (3 moles, γ=53\gamma = \frac{5}{3}γ=35​)

Per mole,

Cv1=R53−1=R23=3R2C_{v1} = \frac{R}{\frac{5}{3}-1} = \frac{R}{\frac{2}{3}} = \frac{3R}{2}Cv1​=35​−1R​=32​R​=23R​ Cp1=Cv1+R=3R2+R=5R2C_{p1} = C_{v1}+R = \frac{3R}{2}+R = \frac{5R}{2}Cp1​=Cv1​+R=23R​+R=25R​

For 333 moles,

CV1total=3⋅3R2=9R2C_{V1}^{\text{total}} = 3\cdot \frac{3R}{2} = \frac{9R}{2}CV1total​=3⋅23R​=29R​ CP1total=3⋅5R2=15R2C_{P1}^{\text{total}} = 3\cdot \frac{5R}{2} = \frac{15R}{2}CP1total​=3⋅25R​=215R​
  1. Diatomic gas (2 moles, γ=75\gamma = \frac{7}{5}γ=57​)

Per mole,

Cv2=R75−1=R25=5R2C_{v2} = \frac{R}{\frac{7}{5}-1} = \frac{R}{\frac{2}{5}} = \frac{5R}{2}Cv2​=57​−1R​=52​R​=25R​ Cp2=Cv2+R=5R2+R=7R2C_{p2} = C_{v2}+R = \frac{5R}{2}+R = \frac{7R}{2}Cp2​=Cv2​+R=25R​+R=27R​

For 222 moles,

CV2total=2⋅5R2=5RC_{V2}^{\text{total}} = 2\cdot \frac{5R}{2} = 5RCV2total​=2⋅25R​=5R CP2total=2⋅7R2=7RC_{P2}^{\text{total}} = 2\cdot \frac{7R}{2} = 7RCP2total​=2⋅27R​=7R
  1. Total heat capacities of the mixture
CV=9R2+5R=9R2+10R2=19R2C_V = \frac{9R}{2} + 5R = \frac{9R}{2} + \frac{10R}{2} = \frac{19R}{2}CV​=29R​+5R=29R​+210R​=219R​ CP=15R2+7R=15R2+14R2=29R2C_P = \frac{15R}{2} + 7R = \frac{15R}{2} + \frac{14R}{2} = \frac{29R}{2}CP​=215R​+7R=215R​+214R​=229R​

Therefore, the adiabatic exponent of the mixture is

γmix=CPCV=29R219R2=2919≈1.526\gamma_{\text{mix}} = \frac{C_P}{C_V} = \frac{\frac{29R}{2}}{\frac{19R}{2}} = \frac{29}{19} \approx 1.526γmix​=CV​CP​​=219R​229R​​=1929​≈1.526
  1. Match with options
γmix≈1.52\gamma_{\text{mix}} \approx 1.52γmix​≈1.52

So the correct option is B.


  1. Comparison with stored answer

Stored correct answer: B
Derived answer: B
They match.

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