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Heat and Thermodynamics question

2019 · 9 Jan · Shift 2 · Q61
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Heat and Thermodynamics question

2019 · 9 Jan · Shift 2 · Q61

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27oC. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled, is about : [Take R = 8.3 J/K mole]
  1. A
    0.9 kJ
  2. B
    6 kJ
  3. C
    10 kJ
  4. D
    14 kJ
View written solutionFree

Correct answer: C

  1. Given data
  • Mass of nitrogen gas: m=15 gm = 15\,\text{g}m=15g
  • Initial temperature: T1=27∘C=300 KT_1 = 27^\circ\text{C} = 300\,\text{K}T1​=27∘C=300K
  • Gas: nitrogen N2N_2N2​, so molar mass M=28 g mol−1M = 28\,\text{g mol}^{-1}M=28g mol−1
  • Gas is enclosed in a vessel ⇒\Rightarrow⇒ volume is constant
  • Gas constant: R=8.3 J mol−1K−1R = 8.3\,\text{J mol}^{-1}\text{K}^{-1}R=8.3J mol−1K−1
  1. Relation between rms speed and temperature

For an ideal gas,

vrms∝Tv_{\text{rms}} \propto \sqrt{T}vrms​∝T​

If rms speed is doubled,

v2v1=2=T2T1\frac{v_2}{v_1} = 2 = \sqrt{\frac{T_2}{T_1}}v1​v2​​=2=T1​T2​​​

Squaring both sides,

T2T1=4\frac{T_2}{T_1} = 4T1​T2​​=4

So,

T2=4T1=4×300=1200 KT_2 = 4T_1 = 4 \times 300 = 1200\,\text{K}T2​=4T1​=4×300=1200K

Thus,

ΔT=T2−T1=1200−300=900 K\Delta T = T_2 - T_1 = 1200 - 300 = 900\,\text{K}ΔT=T2​−T1​=1200−300=900K
  1. Find number of moles
n=mM=1528 moln = \frac{m}{M} = \frac{15}{28}\,\text{mol}n=Mm​=2815​mol
  1. Heat supplied at constant volume

Since the gas is enclosed in a vessel, heating occurs at constant volume.

For diatomic gas N2N_2N2​ (neglecting vibrational modes),

CV=5R2C_V = \frac{5R}{2}CV​=25R​

So,

Q=nCVΔTQ = n C_V \Delta TQ=nCV​ΔT

Substitute values:

Q=1528⋅52⋅8.3⋅900Q = \frac{15}{28} \cdot \frac{5}{2} \cdot 8.3 \cdot 900Q=2815​⋅25​⋅8.3⋅900

Now calculate:

1528≈0.5357\frac{15}{28} \approx 0.53572815​≈0.5357 52⋅8.3=20.75\frac{5}{2} \cdot 8.3 = 20.7525​⋅8.3=20.75 20.75⋅900=1867520.75 \cdot 900 = 1867520.75⋅900=18675 Q≈0.5357×18675≈10004 JQ \approx 0.5357 \times 18675 \approx 10004\,\text{J}Q≈0.5357×18675≈10004J

Thus,

Q≈1.0×104 J=10 kJQ \approx 1.0 \times 10^4\,\text{J} = 10\,\text{kJ}Q≈1.0×104J=10kJ
  1. Option check
  • A: 0.9 kJ0.9\,\text{kJ}0.9kJ — incorrect
  • B: 6 kJ6\,\text{kJ}6kJ — incorrect
  • C: 10 kJ10\,\text{kJ}10kJ — correct
  • D: 14 kJ14\,\text{kJ}14kJ — incorrect

Therefore, the correct option is C.

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