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Heat and Thermodynamics question

2021 · 1 Sep · Shift 2 · Q66
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Heat and Thermodynamics question

2021 · 1 Sep · Shift 2 · Q66

JEE MainPhysicsHeat and ThermodynamicsNumerical+4 / −1
The temperature of 3.00 mol of an ideal diatomic gas is increased by 40.0 ∘^\circ∘ C without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of workdone by the gas is x10{x \over {10}}10x​. Then the value of x (round off to the nearest integer) is ‾\underline{\hspace{2cm}}​. (Given R = 8.31 J mol −-− 1 K −-− 1)
Numerical answer
View written solutionFree

Correct answer: 25

  1. Identify degrees of freedom

For an ideal diatomic gas with rotation allowed but vibration neglected:

  • Translational degrees of freedom =3=3=3
  • Rotational degrees of freedom =2=2=2

So total degrees of freedom: f=5f=5f=5

Hence, internal energy of the gas is U=f2nRT=52nRTU=\frac{f}{2}nRT=\frac{5}{2}nRTU=2f​nRT=25​nRT

Therefore, change in internal energy is ΔU=52nRΔT\Delta U=\frac{5}{2}nR\Delta TΔU=25​nRΔT

  1. Given data
  • Number of moles: n=3.00n=3.00n=3.00
  • Temperature increase: ΔT=40.0∘C=40.0 K\Delta T=40.0^\circ\text{C}=40.0\,\text{K}ΔT=40.0∘C=40.0K
  1. Calculate change in internal energy

ΔU=52(3)(8.31)(40)\Delta U=\frac{5}{2}(3)(8.31)(40)ΔU=25​(3)(8.31)(40)

ΔU=2.5×3×8.31×40\Delta U=2.5\times 3\times 8.31\times 40ΔU=2.5×3×8.31×40

ΔU=2493 J\Delta U=2493\,\text{J}ΔU=2493J

  1. Work done in isobaric process

At constant pressure, work done by an ideal gas is W=nRΔTW=nR\Delta TW=nRΔT

So, W=(3)(8.31)(40)=997.2 JW=(3)(8.31)(40)=997.2\,\text{J}W=(3)(8.31)(40)=997.2J

  1. Find the ratio

ΔUW=2493997.2=2.5=2510\frac{\Delta U}{W}=\frac{2493}{997.2}=2.5=\frac{25}{10}WΔU​=997.22493​=2.5=1025​

Thus, ΔUW=x10\frac{\Delta U}{W}=\frac{x}{10}WΔU​=10x​

So, x=25x=25x=25

  1. Comparison with stored answer

Stored correct answer = 252525

Our derived answer also gives 252525.

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