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Heat and Thermodynamics question

2019 · 12 Jan · Shift 1 · Q65
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Heat and Thermodynamics question

2019 · 12 Jan · Shift 1 · Q65

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
An ideal gas occupies a volume of 2m3 at a pressure of 3 ×\times× 106 Pa. The energy of the gas is :
  1. A
    6 ×\times× 104 J
  2. B
    9 ×\times× 106 J
  3. C
    3 ×\times× 102 J
  4. D
    108 J
View written solutionFree

Correct answer: B

  1. Given data
  • Pressure: P=3×106 PaP = 3 \times 10^6\ \text{Pa}P=3×106 Pa
  • Volume: V=2 m3V = 2\ \text{m}^3V=2 m3

We are asked to find the energy of the ideal gas.

  1. Use the relation for internal energy

For an ideal gas, the internal energy depends on temperature. In many such MCQ problems, “energy of gas” is taken as

U=32PVU = \frac{3}{2}PVU=23​PV

(for a monoatomic ideal gas).

  1. Substitute the values

First calculate PVPVPV:

PV=(3×106)(2)=6×106 JPV = (3 \times 10^6)(2) = 6 \times 10^6\ \text{J}PV=(3×106)(2)=6×106 J

Now,

U=32×6×106U = \frac{3}{2} \times 6 \times 10^6U=23​×6×106

U=9×106 JU = 9 \times 10^6\ \text{J}U=9×106 J

  1. Match with options
  • A: 6×1046 \times 10^46×104 J
  • B: 9×1069 \times 10^69×106 J
  • C: 3×1023 \times 10^23×102 J
  • D: 10810^8108 J

So the correct option is:

B\boxed{B}B​

  1. Comparison with stored answer

Stored correct answer: B

My derived answer: B

They match.

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