A vessel contains 14 g of nitrogen gas at a temperature of 27°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 mole1 k1.

Option 1 - <p><strong>&nbsp;2229 J</strong></p>
Option 2 - <p><strong>5616 J</strong></p>
Option 3 - <p><strong>9360 J</strong></p>
Option 4 - <p><strong>13, 104 J</strong></p>
3 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
V
7 months ago
Correct Option - 3
Detailed Solution:

 mN2=14g, T=27°C=300k

To double the Vrms Tf = 4Ti = 4 * 300 = 1200 k.

Q=nCvΔT

=5*225R=1125R=1125*3.32

= 9360 J.

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The average kinetic energy (per molecule)of any (ideal) gas (be it monoatomic like Argon, diatomic like oxygen or polyatomic) is always equal to 32KBT. If depends only on temperature, and is independent of the nature of the gas. So, Answer is 1 : 1.

Degree of freedom, f = 8

CV=82R=4RandCP=4R+R=5R

So, H = ΔU+W

at constant pressure, W = PΔv

150 = nRΔT

So,  H=nCVΔT=n5RΔT=5 (nRΔT)

= 5 × 150

= 750 J

x ¯ (mean free path) = 1 2 π d 2 n v  

n v = n N A v , n → No. of moles in volume v NA ® Avogadro’s Number

n v = P R T

x ¯ v 1 ρ ( d e n s i t y & x ¯ α T a t c o n s t a n t P )

ρ x ¯ | x ¯ T                           

The motion of the gas molecules freezes at 0K not 0° C

Average kinetic Energy per molecule

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Physics NCERT Exemplar Solutions Class 12th Chapter Eleven 2025

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