Question

Compute the temperature at which the r.m.s. The speed of nitrogen molecules is 832 m/s. [Universal gas constant $ R = 8320J/kmole \cdot K $ , molecular weight of nitrogen $ = 28 $ ].

Answer 1

Hint: The root mean square speed is related to the absolute temperature via the expression $ {v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ . We need to make the absolute temperature subject of the formula and from there we can calculate the answer.

Formula used: In this solution we will be using the following formula;
$ \Rightarrow {v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ where $ {v_{rms}} $ is the root mean square (r.m.s) speed, $ R $ is the universal gas constant, is the absolute temperature and M is the molecular weight of the gas.

Complete step by step answer
In general, the temperature of a body is a measure of the average kinetic energy of the body. The higher the temperature, the faster the average kinetic energy and the lower the temperature, the lower the kinetic energy. Hence, from the temperature of a substance the average velocity can be known. This average velocity is given as the root mean square velocity. In vice versa, the temperature of a substance can also be calculated from its average velocity.
Solving the question, $ {v_{rms}} $ is generally given by $ {v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ where $ R $ is the universal gas constant, $ T $ is the absolute temperature and $ M $ is the molecular weight of the gas.
From the above equation, we can make $ T $ subject of the formula,by squaring both sides, we have
 $ \Rightarrow {\left( {{v_{rms}}} \right)^2} = \dfrac{{3RT}}{M} $
Then we multiply both sides by $ M $ and divide by $ 3R $ , we have that
 $ \Rightarrow T = M\dfrac{{v_{rms}^2}}{{3R}} $
By inserting known values give in the question we get,
 $ \Rightarrow T = \dfrac{{\left( {28 \times {{10}^{ - 3}}} \right){{832}^2}}}{{3\left( {8.32} \right)}} $
By converting all units to SI unit we get,
 $ \Rightarrow T = 776.53K $ .

Note
To clarify any confusion, the unit of molecular weight (even when written without a unit) is usually in g/mol. Hence to convert to SI unit i.e. kg/mol, we have to multiply by $ {10^{ - 3}} $ as in
 $ \Rightarrow 28\dfrac{g}{{mol}} = 28\dfrac{g}{{mol}} \times \dfrac{{1kg}}{{1000g}} = 28 \times {10^{ - 3}}kg $
Similarly, the universal gas constant was given in $ J/kmole \cdot K $ , but SI is given in $ J/mole \cdot K $
Hence, to convert we must divide by $ {10^3} $ as in
 $ \Rightarrow 8320\dfrac{J}{{kmole \cdot K}} = 8320\dfrac{J}{{{{10}^3}mole \cdot K}} = 8.32J/mole \cdot K $ .