# At what temperature, the rms velocity of gas molecules would be double of its value at NTP, if pressure is remaining constant?

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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

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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 (Ek 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 Vrms 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.