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Heat and Thermodynamics question

2020 · 7 Jan · Shift 1 · Q57
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Heat and Thermodynamics question

2020 · 7 Jan · Shift 1 · Q57

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
Two moles of an ideal gas with CPCV=53{{{C_P}} \over {{C_V}}} = {5 \over 3}CV​CP​​=35​ are mixed with 3 moles of another ideal gas with CPCV=43{{{C_P}} \over {{C_V}}} = {4 \over 3}CV​CP​​=34​. The value of CPCV{{{C_P}} \over {{C_V}}}CV​CP​​ for the mixture is :
  1. A
    1.50
  2. B
    1.45
  3. C
    1.47
  4. D
    1.42
View written solutionFree

Correct answer: D

  1. Given data
  • Gas 1: n1=2n_1 = 2n1​=2 moles, γ1=CP1CV1=53\gamma_1 = \dfrac{C_{P1}}{C_{V1}} = \dfrac{5}{3}γ1​=CV1​CP1​​=35​
  • Gas 2: n2=3n_2 = 3n2​=3 moles, γ2=CP2CV2=43\gamma_2 = \dfrac{C_{P2}}{C_{V2}} = \dfrac{4}{3}γ2​=CV2​CP2​​=34​

We need the ratio for the mixture:

γmix=CP,mixCV,mix\gamma_{\text{mix}} = \frac{C_{P,\text{mix}}}{C_{V,\text{mix}}}γmix​=CV,mix​CP,mix​​

For ideal gases:

CP−CV=RC_P - C_V = RCP​−CV​=R

and

γ=CPCV\gamma = \frac{C_P}{C_V}γ=CV​CP​​

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. Find molar heat capacities of gas 1

For gas 1, γ1=53\gamma_1 = \dfrac{5}{3}γ1​=35​:

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 2 moles:

CV1,total=2⋅3R2=3RC_{V1,\text{total}} = 2\cdot \frac{3R}{2} = 3RCV1,total​=2⋅23R​=3R CP1,total=2⋅5R2=5RC_{P1,\text{total}} = 2\cdot \frac{5R}{2} = 5RCP1,total​=2⋅25R​=5R
  1. Find molar heat capacities of gas 2

For gas 2, γ2=43\gamma_2 = \dfrac{4}{3}γ2​=34​:

CV2=R43−1=R13=3RC_{V2} = \frac{R}{\frac{4}{3}-1} = \frac{R}{\frac{1}{3}} = 3RCV2​=34​−1R​=31​R​=3R CP2=CV2+R=3R+R=4RC_{P2} = C_{V2} + R = 3R + R = 4RCP2​=CV2​+R=3R+R=4R

For 3 moles:

CV2,total=3⋅3R=9RC_{V2,\text{total}} = 3\cdot 3R = 9RCV2,total​=3⋅3R=9R CP2,total=3⋅4R=12RC_{P2,\text{total}} = 3\cdot 4R = 12RCP2,total​=3⋅4R=12R
  1. Add heat capacities for the mixture
CV,mix=3R+9R=12RC_{V,\text{mix}} = 3R + 9R = 12RCV,mix​=3R+9R=12R CP,mix=5R+12R=17RC_{P,\text{mix}} = 5R + 12R = 17RCP,mix​=5R+12R=17R

Hence,

γmix=CP,mixCV,mix=17R12R=1712\gamma_{\text{mix}} = \frac{C_{P,\text{mix}}}{C_{V,\text{mix}}} = \frac{17R}{12R} = \frac{17}{12}γmix​=CV,mix​CP,mix​​=12R17R​=1217​ γmix≈1.4167\gamma_{\text{mix}} \approx 1.4167γmix​≈1.4167
  1. Match with options
1.4167≈1.421.4167 \approx 1.421.4167≈1.42

So the correct option is:

D: 1.42

  1. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

They agree.

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