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Heat and Thermodynamics question

2009 · Shift 0 · Q70
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Heat and Thermodynamics question

2009 · Shift 0 · Q70

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
One kgkgkg of a diatomic gas is at a pressure of 8×104 N/m2.8 \times {10^4}\,N/{m^2}.8×104N/m2. The density of the gas is 4kg/m34kg/{m^3}4kg/m3. What is the energy of the gas due to its thermal motion ?
  1. A
    5×104 J5 \times {10^4}\,J5×104J
  2. B
    6×104 J6 \times {10^4}\,J6×104J
  3. C
    7×104 J7 \times {10^4}\,J7×104J
  4. D
    3×104 J3 \times {10^4}\,J3×104J
View written solutionFree

Correct answer: A

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

The gas is diatomic, so for ordinary temperatures its degrees of freedom are f=5f=5f=5.

  1. Find the volume of the gas

Using

ρ=mV\rho = \frac{m}{V}ρ=Vm​

so

V=mρ=14=0.25 m3V = \frac{m}{\rho} = \frac{1}{4} = 0.25\,\text{m}^3V=ρm​=41​=0.25m3
  1. Use the formula for thermal (internal) energy

For an ideal gas,

U=f2PVU = \frac{f}{2}PVU=2f​PV

For a diatomic gas,

U=52PVU = \frac{5}{2}PVU=25​PV

Now,

PV=(8×104)(0.25)=2×104 JPV = (8 \times 10^4)(0.25) = 2 \times 10^4\,\text{J}PV=(8×104)(0.25)=2×104J

Therefore,

U=52(2×104)=5×104 JU = \frac{5}{2}(2 \times 10^4) = 5 \times 10^4\,\text{J}U=25​(2×104)=5×104J
  1. Check options
  • A: 5×104 J5 \times 10^4\,\text{J}5×104J ✅
  • B: 6×104 J6 \times 10^4\,\text{J}6×104J
  • C: 7×104 J7 \times 10^4\,\text{J}7×104J
  • D: 3×104 J3 \times 10^4\,\text{J}3×104J

So the correct option is A.

  1. Comparison with stored answer

Stored correct answer = A

Our derived answer also = A, so they agree.

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