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Heat and Thermodynamics question

2013 · Shift 1 · Q49
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Heat and Thermodynamics question

2013 · Shift 1 · Q49

JEE AdvancedPhysicsHeat and ThermodynamicsMCQ+2 / −0.5
Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is
  1. A
    1 : 4
  2. B
    1 : 2
  3. C
    6 : 9
  4. D
    8 : 9
View written solutionFree

Correct answer: D

Step-by-step Solution

  1. Recall the Ideal Gas Law: The ideal gas equation is given by PV=nRTPV = nRTPV=nRT, where PPP is the pressure, VVV is the volume, nnn is the number of moles, RRR is the universal gas constant, and TTT is the absolute temperature.

  2. Express the Ideal Gas Law in terms of density: The number of moles nnn can be written as the ratio of the total mass of the gas mmm to its molar mass MMM: n=m/Mn = m/Mn=m/M. Substituting this into the ideal gas equation gives: PV=mMRTPV = \frac{m}{M}RTPV=Mm​RT Density, ρ\rhoρ, is defined as mass per unit volume, ρ=m/V\rho = m/Vρ=m/V. We can rearrange the equation to solve for density: mV=PMRT\frac{m}{V} = \frac{PM}{RT}Vm​=RTPM​ So, the density of an ideal gas is given by: ρ=PMRT\rho = \frac{PM}{RT}ρ=RTPM​

  3. Set up the ratio of densities for the two gases: Let the two gases be Gas 1 and Gas 2. Their properties will be denoted by subscripts 1 and 2, respectively. The density of Gas 1 is ρ1=P1M1RT\rho_1 = \frac{P_1 M_1}{RT}ρ1​=RTP1​M1​​. The density of Gas 2 is ρ2=P2M2RT\rho_2 = \frac{P_2 M_2}{RT}ρ2​=RTP2​M2​​. Note that both gases are in the same vessel at a constant temperature, so the volume VVV, temperature TTT, and the gas constant RRR are the same for both.

    The ratio of their densities is: ρ1ρ2=P1M1RTP2M2RT=P1M1P2M2\frac{\rho_1}{\rho_2} = \frac{\frac{P_1 M_1}{RT}}{\frac{P_2 M_2}{RT}} = \frac{P_1 M_1}{P_2 M_2}ρ2​ρ1​​=RTP2​M2​​RTP1​M1​​​=P2​M2​P1​M1​​ This can be rewritten as: ρ1ρ2=(P1P2)(M1M2)\frac{\rho_1}{\rho_2} = \left(\frac{P_1}{P_2}\right) \left(\frac{M_1}{M_2}\right)ρ2​ρ1​​=(P2​P1​​)(M2​M1​​)

  4. Substitute the given values into the ratio equation: From the problem statement, we have:

    • The ratio of partial pressures: P1P2=43\frac{P_1}{P_2} = \frac{4}{3}P2​P1​​=34​
    • The ratio of atomic masses (and thus molar masses for monoatomic gases): M1M2=23\frac{M_1}{M_2} = \frac{2}{3}M2​M1​​=32​

    Substituting these values into the density ratio equation: ρ1ρ2=(43)(23)\frac{\rho_1}{\rho_2} = \left(\frac{4}{3}\right) \left(\frac{2}{3}\right)ρ2​ρ1​​=(34​)(32​)

  5. Calculate the final ratio of densities: ρ1ρ2=4×23×3=89\frac{\rho_1}{\rho_2} = \frac{4 \times 2}{3 \times 3} = \frac{8}{9}ρ2​ρ1​​=3×34×2​=98​ So, the ratio of the densities of the two gases is 8 : 9.

  6. Compare with the given options: The calculated ratio 8 : 9 matches option D.

    A: 1 : 4 B: 1 : 2 C: 6 : 9 (which simplifies to 2 : 3) D: 8 : 9

    The correct option is D.

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