Question

At what temperature, the rms velocity of gas molecules would be double of its value at NTP, if pressure is remaining constant?
1. 819°C 
2. 819 K
3. 846 K
4. 546°C 

Answer 1

Correct Answer - Option 1 : 819°C 

CONCEPT:

Root mean square (R.M.S.) velocity.

• we know how to relate temperature and kinetic energy, we can relate temperature to the velocity of gas molecules. Note since these are distributions the values (E_k or velocity) that we are talking about are always averages 

\(E_k = \frac{3}{2}RT\)

\(E_k = \frac{1}{2}mv^{2}\)

• setting these two equal and solving for the average square velocity we get.

\(v^{2} = \frac{3RT}{m}\)

• The root mean square velocity or V_rms is the square root of the average square velocity and is

\(R.M.S = \sqrt{\frac{3RT}{M}}\)

where R = gas constant, T = temperature (in K), M = molar mass of the gas

​​EXPLANATIONS:​

• From the gas equation, we know

\(c = \sqrt{\frac{3RT}{m}}\)

• In given condition, 

\(\frac{C_T}{C_o}= \sqrt{\frac{T}{T_o}} => \frac{2C_o}{C_o} = \sqrt{\frac{T}{T_o}}\)

\(=> 4 = \frac{T}{T_o}\)

\(=>T = 4T_o = 4 \times 273 = 1092K = 819^0 C\)

• hence option 1 is the correct answer.