JEE Challenger
More from Thermodynamics
PhyMedAdv 2023 • Paper 1

Internal Energy Change of Monatomic and Diatomic Gas Mixture

A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas (γ=5/3\gamma = 5/3) and one mole of an ideal diatomic gas (γ=7/5\gamma = 7/5). Here, γ\gamma is the ratio of the specific heats at constant pressure and constant volume of an ideal gas. The gas mixture does a work of 66 Joule66\text{ Joule} when heated at constant pressure. The change in its internal energy is ________ Joule\text{Joule}.

Official Numerical Answer121

Topics & Concepts

ThermodynamicsFirst Law of ThermodynamicsKinetic Theory of Gases

Step-by-Step Solution

To find the change in the internal energy of the gas mixture, we use the principles of thermodynamics for ideal gases.

1. Given Data:

  • Number of moles of monatomic gas, n1=2n_1 = 2
  • Molar heat capacity at constant volume for the monatomic gas: Cv1=32RC_{v1} = \frac{3}{2}R
  • Number of moles of diatomic gas, n2=1n_2 = 1
  • Molar heat capacity at constant volume for the diatomic gas: Cv2=52RC_{v2} = \frac{5}{2}R
  • Total number of moles, n=n1+n2=2+1=3 molesn = n_1 + n_2 = 2 + 1 = 3\text{ moles}
  • Work done by the mixture at constant pressure, W=66 JouleW = 66\text{ Joule}

2. Relation for Work Done at Constant Pressure:

For an ideal gas undergoing an isobaric (constant pressure) process, the work done WW is given by: W=PΔV=nRΔTW = P \Delta V = n R \Delta T

Substituting the given values: (n1+n2)RΔT=66(n_1 + n_2) R \Delta T = 66 3RΔT=663 R \Delta T = 66 RΔT=22 JouleR \Delta T = 22\text{ Joule}

3. Calculation of Change in Internal Energy (ΔU\Delta U):

The change in internal energy of the gas mixture is the sum of the changes in internal energy of each component: ΔU=ΔU1+ΔU2=n1Cv1ΔT+n2Cv2ΔT\Delta U = \Delta U_1 + \Delta U_2 = n_1 C_{v1} \Delta T + n_2 C_{v2} \Delta T

Substituting the values of n1,Cv1,n2,n_1, C_{v1}, n_2, and Cv2C_{v2}: ΔU=2(32R)ΔT+1(52R)ΔT\Delta U = 2 \left(\frac{3}{2}R\right) \Delta T + 1 \left(\frac{5}{2}R\right) \Delta T ΔU=3RΔT+52RΔT\Delta U = 3R \Delta T + \frac{5}{2}R \Delta T ΔU=112RΔT\Delta U = \frac{11}{2} R \Delta T

Substituting RΔT=22 JouleR \Delta T = 22\text{ Joule}: ΔU=112×22=11×11=121 Joule\Delta U = \frac{11}{2} \times 22 = 11 \times 11 = 121\text{ Joule}

Final Answer:

The change in its internal energy is 121121 Joule.

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