Question

At what temperature is the ${{v}_{rms}}$ of molecules of a given gas is four times the ${{v}_{rms}}$ at standard temperature?

A. 3468 K
B. 6348 K
C. 4368 K
D. 8436 K

Answer 1

Using the formula for calculating the RMS voltage of the gas, this question can be solved. Firstly find the RMS voltage of the gas at the standard temperature. Then, consider the RMS voltage to be 4 times the RMS voltage of gas at T temperature. Finally, divide these values of the RMS voltage to find the temperature of the gas.
Formula used:
${{v}_{rms}}=\sqrt{\dfrac{3RT}{M}}$

Complete answer:
The formula that is used to find the RMS voltage of the gas is given as follows.
${{v}_{rms}}=\sqrt{\dfrac{3RT}{M}}$
Where R is the gas constant, T is the temperature and M is the mass of the gas molecules.
We need to consider the two cases.

Case I: The RMS voltage of the gas at a standard temperature.
${{v}_{rms}}_{1}=\sqrt{\dfrac{3R{{T}_{1}}}{M}}$
Case II: The 4 times the RMS voltage of the gas at a standard temperature.
${{v}_{rms}}_{2}=\sqrt{\dfrac{3R{{T}_{2}}}{M}}$
We can remove the constant parameters like the gas constant and the mass of the gas molecules, as both of these will remain the same.
So, we have,
$\dfrac{{{v}_{rms}}_{1}}{{{v}_{rms}}_{2}}=\sqrt{\dfrac{{{T}_{1}}}{{{T}_{2}}}}$
From the data, we have the data, the rms voltage of a gas is 4 times the rms voltage of the gas at a standard temperature. That is, ${{v}_{rms}}_{2}=4{{v}_{rms}}_{1}$.
The standard temperature of the RMS voltage of the gas is, 273 K.
Substitute the values of the given rms voltage and the standard temperature in the above equation. Thus, we get,
$\dfrac{{{v}_{rms}}_{1}}{4{{v}_{rms}}_{1}}=\sqrt{\dfrac{273}{{{T}_{2}}}}$
Continue the further calculation.

& \dfrac{1}{4}=\sqrt{\dfrac{273}{{{T}_{2}}}} \\\ & \Rightarrow \dfrac{1}{16}=\dfrac{273}{{{T}_{2}}} \\\ \end{aligned}$$ Now find the value of the temperature. $$\begin{aligned} & {{T}_{2}}=273\times 16 \\\ & \Rightarrow {{T}_{2}}=4368\,K \\\ \end{aligned}$$ At the temperature of 4368 K the $${{v}_{rms}}$$ of molecules of a given gas is four times the $${{v}_{rms}}$$ at standard temperature. **So, the correct answer is “Option A”.** **Note:** In this case, the temperature value is asked. Even by giving the values of the temperatures, the voltage of the gas can be asked. Here in the options, the temperature in the units of Kelvin is given. Even, by giving the unit of the temperature in Celsius, they can ask to find the value. In such a case, to convert the Kelvin into Celsius, add the value of 273 to the temperature value given in Celsius.