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Heat and Thermodynamics question

2022 · 26 Jul · Shift 1 · Q56
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  5. /2022 · 26 Jul · Shift 1 · Q56

Heat and Thermodynamics question

2022 · 26 Jul · Shift 1 · Q56

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
7 mol of a certain monoatomic ideal gas undergoes a temperature increase of 40 K40 \mathrm{~K}40 K at constant pressure. The increase in the internal energy of the gas in this process is : (Given R=8.3 JK−1 mol−1\mathrm{R}=8.3 \,\mathrm{JK}^{-1} \mathrm{~mol}^{-1}R=8.3JK−1 mol−1 )
  1. A
    5810 J
  2. B
    3486 J
  3. C
    11620 J
  4. D
    6972 J
View written solutionFree

Correct answer: B

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

For an ideal gas,

ΔU=nCVΔT\Delta U = n C_V \Delta TΔU=nCV​ΔT

For a monoatomic ideal gas,

CV=32RC_V = \frac{3}{2}RCV​=23​R

So,

ΔU=n(32R)ΔT\Delta U = n\left(\frac{3}{2}R\right)\Delta TΔU=n(23​R)ΔT
  1. Substitute the given values

Given:

  • n=7n = 7n=7 mol
  • ΔT=40 K\Delta T = 40\,\text{K}ΔT=40K
  • R=8.3 J mol−1K−1R = 8.3\,\text{J mol}^{-1}\text{K}^{-1}R=8.3J mol−1K−1

Thus,

ΔU=7×32×8.3×40\Delta U = 7 \times \frac{3}{2} \times 8.3 \times 40ΔU=7×23​×8.3×40
  1. Calculate step by step

First,

32×8.3=12.45\frac{3}{2} \times 8.3 = 12.4523​×8.3=12.45

Then,

12.45×40=49812.45 \times 40 = 49812.45×40=498

Finally,

498×7=3486498 \times 7 = 3486498×7=3486

So,

ΔU=3486 J\Delta U = 3486\,\text{J}ΔU=3486J
  1. Match with the options

3486 J3486\,\text{J}3486J corresponds to Option B.

  1. Comparison with stored correct answer

Stored correct answer: B

This matches our derived answer.

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