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Heat and Thermodynamics question

2019 · 10 Jan · Shift 2 · Q64
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Heat and Thermodynamics question

2019 · 10 Jan · Shift 2 · Q64

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
Two kg of a monoatomic gas is at a pressure of 4 ×\times× 104 N/m2. The density of the gas is 8 kg/m3. What is the order of energy of the gas due to its thermal motion ?
  1. A
    104 J
  2. B
    103 J
  3. C
    105 J
  4. D
    106 J
View written solutionFree

Correct answer: A

  1. Given data
  • Mass of gas: m=2 kgm = 2\,\text{kg}m=2kg
  • Pressure: P=4×104 N/m2P = 4 \times 10^4\,\text{N/m}^2P=4×104N/m2
  • Density: ρ=8 kg/m3\rho = 8\,\text{kg/m}^3ρ=8kg/m3
  • Gas is monoatomic

We need the energy due to thermal motion, i.e. the internal thermal energy of a monoatomic ideal gas.

  1. Find the volume of the gas

Using

ρ=mV\rho = \frac{m}{V}ρ=Vm​

so

V=mρ=28=0.25 m3V = \frac{m}{\rho} = \frac{2}{8} = 0.25\,\text{m}^3V=ρm​=82​=0.25m3
  1. Use internal energy formula for monoatomic ideal gas

For a monoatomic gas,

U=32PVU = \frac{3}{2}PVU=23​PV

Now,

PV=(4×104)(0.25)=104PV = (4 \times 10^4)(0.25) = 10^4PV=(4×104)(0.25)=104

Therefore,

U=32×104=1.5×104 JU = \frac{3}{2} \times 10^4 = 1.5 \times 10^4\,\text{J}U=23​×104=1.5×104J
  1. Determine the order of magnitude

The value is

1.5×104 J1.5 \times 10^4\,\text{J}1.5×104J

So its order is

104 J10^4\,\text{J}104J
  1. Check options
  • A: 104 J10^4\,\text{J}104J ✅
  • B: 103 J10^3\,\text{J}103J ❌
  • C: 105 J10^5\,\text{J}105J ❌
  • D: 106 J10^6\,\text{J}106J ❌

Hence, the correct option is A.

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