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Heat and Thermodynamics question

2019 · 11 Jan · Shift 1 · Q69
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Heat and Thermodynamics question

2019 · 11 Jan · Shift 1 · Q69

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. considering only translational and rotational modes, the total internal energy of the system is :
  1. A
    12 RT
  2. B
    20 RT
  3. C
    4 RT
  4. D
    15 RT
View written solutionFree

Correct answer: D

  1. Use internal energy formula for ideal gases

For an ideal gas, U=nCVTU = n C_V TU=nCV​T

If we count only translational and rotational degrees of freedom, then:

  • For a monatomic gas like argon, degrees of freedom f=3f=3f=3 U=f2nRT=32nRTU = \frac{f}{2}nRT = \frac{3}{2}nRTU=2f​nRT=23​nRT
  • For a diatomic gas like oxygen, ignoring vibration and counting translational + rotational only, f=5f=5f=5 U=52nRTU = \frac{5}{2}nRTU=25​nRT
  1. Internal energy of oxygen

Given 333 moles of O2O_2O2​: UO2=52×3RT=152RTU_{O_2} = \frac{5}{2}\times 3RT = \frac{15}{2}RTUO2​​=25​×3RT=215​RT

  1. Internal energy of argon

Given 555 moles of Ar: UAr=32×5RT=152RTU_{Ar} = \frac{3}{2}\times 5RT = \frac{15}{2}RTUAr​=23​×5RT=215​RT

  1. Total internal energy

Utotal=UO2+UArU_{total} = U_{O_2} + U_{Ar}Utotal​=UO2​​+UAr​ Utotal=152RT+152RT=15RTU_{total} = \frac{15}{2}RT + \frac{15}{2}RT = 15RTUtotal​=215​RT+215​RT=15RT

  1. Match with options

Thus the correct option is: 15RT\boxed{15RT}15RT​ which is Option D.

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