Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Help me solve this A 15g mass of nitrogen gas is enclosed in a vessel at a temperature 270C. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled , is about: [ Take R = 8.3J/K mole ]

A 15g mass of nitrogen gas is enclosed in a vessel at a temperature 270C. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled , is about:

[ Take R = 8.3J/K mole ]

  • Option 1)

     

    10kJ

  • Option 2)

     

    6kJ

  • Option 3)

     

    14 kJ

  • Option 4)

     

    0.9 kJ

Answers (1)

best_answer

 

Root mean square velocity -

V_{rms}= \sqrt{\frac{3RT}{M}}

= \sqrt{\frac{3P}{\rho }}
 

- wherein

R = Universal gas constant

M = molar mass

P = pressure due to gas

\rho = density

As gas is closed vessel 

Q=nC_{V}\Delta T

V\propto \sqrt T

RMS velocity is doubled 

=>T_{2}=4T_{1}

\therefore  Q=nC_{V}(4T_{1}-T_{1})

          =nC_{V} \cdot 3T_{1}

         =\frac{15}{28}\times \frac{5}{2}R\times 3\times 300

        =10000J

        =10kJ

 

 


Option 1)

 

10kJ

Option 2)

 

6kJ

Option 3)

 

14 kJ

Option 4)

 

0.9 kJ

Posted by

admin

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: IIT Roorkee’s ECE emerges as top choice after CSE; closing rank rises
anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus

Latest Question