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Heat and Thermodynamics question

2023 · 10 Apr · Shift 2 · Q43
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Heat and Thermodynamics question

2023 · 10 Apr · Shift 2 · Q43

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A gas mixture consists of 2 moles of oxygen and 4 moles of neon at temperature T. Neglecting all vibrational modes, the total internal energy of the system will be,
  1. A
    4RT
  2. B
    16RT
  3. C
    8RT
  4. D
    11RT
View written solutionFree

Correct answer: D

  1. Use the formula for internal energy of an ideal gas mixture

For an ideal gas, internal energy is the sum over all components:

U=∑ni(fi2RT)U = \sum n_i \left(\frac{f_i}{2}RT\right)U=∑ni​(2fi​​RT)

where:

  • nin_ini​ = number of moles of the gas
  • fif_ifi​ = degrees of freedom
  • RRR = gas constant
  • TTT = temperature
  1. Find degrees of freedom of each gas

  • Oxygen (O2O_2O2​) is a diatomic gas. Neglecting vibrational modes, it has:

    • 3 translational
    • 2 rotational

    So, fO2=5f_{O_2} = 5fO2​​=5

  • Neon (Ne) is a monoatomic gas. It has only 3 translational degrees of freedom: fNe=3f_{Ne} = 3fNe​=3

  1. Compute internal energy of oxygen

Given 2 moles of oxygen:

UO2=2⋅52RT=5RTU_{O_2} = 2 \cdot \frac{5}{2}RT = 5RTUO2​​=2⋅25​RT=5RT

  1. Compute internal energy of neon

Given 4 moles of neon:

UNe=4⋅32RT=6RTU_{Ne} = 4 \cdot \frac{3}{2}RT = 6RTUNe​=4⋅23​RT=6RT

  1. Add both contributions

Utotal=UO2+UNeU_{total} = U_{O_2} + U_{Ne}Utotal​=UO2​​+UNe​ Utotal=5RT+6RT=11RTU_{total} = 5RT + 6RT = 11RTUtotal​=5RT+6RT=11RT

  1. Match with the options

U=11RT\boxed{U = 11RT}U=11RT​

So the correct option is D.

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