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Heat and Thermodynamics question

2024 · 29 Jan · Shift 2 · Q76
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  5. /2024 · 29 Jan · Shift 2 · Q76

Heat and Thermodynamics question

2024 · 29 Jan · Shift 2 · Q76

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
NNN moles of a polyatomic gas (f=6)(f=6)(f=6) must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of NNN is :
  1. A
    6
  2. B
    2
  3. C
    4
  4. D
    3
View written solutionFree

Correct answer: C

  1. For an ideal gas, the ratio of specific heats depends on the degrees of freedom fff:
γ=CPCV=f+2f\gamma = \frac{C_P}{C_V} = \frac{f+2}{f}γ=CV​CP​​=ff+2​

Also,

CV=f2Rper moleC_V = \frac{f}{2}R \quad \text{per mole}CV​=2f​Rper mole
  1. Given gases:
  • Polyatomic gas: f=6f=6f=6
  • Monoatomic gas: f=3f=3f=3
  • Required mixture should behave as a diatomic gas, so effective degrees of freedom must be f=5f=5f=5.
  1. Total molar heat capacity at constant volume for the mixture:

For NNN moles of polyatomic gas,

CV1=N(62R)=3NRC_{V1} = N\left(\frac{6}{2}R\right)=3NRCV1​=N(26​R)=3NR

For 222 moles of monoatomic gas,

CV2=2(32R)=3RC_{V2} = 2\left(\frac{3}{2}R\right)=3RCV2​=2(23​R)=3R

So total,

CV(mix)=3NR+3R=3R(N+1)C_V^{(\text{mix})}=3NR+3R=3R(N+1)CV(mix)​=3NR+3R=3R(N+1)
  1. Total moles in the mixture:
N+2N+2N+2

If the mixture behaves like a diatomic gas, then its molar heat capacity at constant volume should be

CV(molar)=52RC_V^{(\text{molar})}=\frac{5}{2}RCV(molar)​=25​R

Hence,

CV(mix)N+2=52R\frac{C_V^{(\text{mix})}}{N+2} = \frac{5}{2}RN+2CV(mix)​​=25​R

Substitute:

3R(N+1)N+2=52R\frac{3R(N+1)}{N+2} = \frac{5}{2}RN+23R(N+1)​=25​R

Cancel RRR:

3(N+1)N+2=52\frac{3(N+1)}{N+2} = \frac{5}{2}N+23(N+1)​=25​
  1. Solve for NNN:
6(N+1)=5(N+2)6(N+1)=5(N+2)6(N+1)=5(N+2) 6N+6=5N+106N+6=5N+106N+6=5N+10 N=4N=4N=4
  1. Checking options:
  • A: 666 ❌
  • B: 222 ❌
  • C: 444 ✅
  • D: 333 ❌

Therefore, the correct answer is Option C.

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