Verified Solution • 2025–26 Session moderate

QJEEM22Phe86f: JEE Main 2022 Physics - Class 11 JEEM22Phe86f Solution: A vessel contains of nitrogen gas at a temperature of . The…

A vessel contains 14 g14 \mathrm{~g} of nitrogen gas at a temperature of 27∘C27^{\circ} \mathrm{C} . The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be :

Take R=8.32 J mol−1 k−1\mathrm{R}=8.32 \mathrm{~J} \mathrm{~mol}^{-1} \,\mathrm{k}^{-1} .

Multiple Choice Options

  • A 2229 J
  • B 5616 J
  • C 9360 J
  • D 13,104 J

Step-by-Step Verified Solution

n = 0.5

T = 300

For vrms to be doubled T' = 4 ×\times 300 = 1200

⇒\Rightarrow Heat transferred

=(0.5)(52)(8.32)(900)= (0.5)\left( {{5 \over 2}} \right)(8.32)(900)

=9360= 9360 J

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