
Nitrogen gas is at \[{300^0}C\] temperature. Find the temperature (in K) at which the rms speed of a \[{H_2}\] molecule would be equal to the rms speed of a nitrogen molecule.
Answer
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Hint: Before we proceed to solving the problem. Let us know about the root mean square velocity (speed). The root mean square velocity is the square root of the mean of squares of the velocity of individual gas molecules.
Formula Used:
The formula for the root mean square speed is,
\[{V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \]
Where, R is universal gas constant, T is temperature and M is molar mass of gas.
Complete step by step solution:
Consider a nitrogen gas at \[{300^0}C\]. We need to find the value of temperature (in K) at which the rms speed of an \[{H_2}\] molecule would be equal to the RMS speed of a nitrogen molecule. For that, we have the formula for finding rms speed of nitrogen as,
\[{\left( {{V_{rms}}} \right)_{{N_2}}} = \sqrt {\dfrac{{3RT}}{M}} \]
We need to convert the temperature into kelvin, that is
\[T = \left( {300 + 273} \right)K\]
\[ \Rightarrow T = 573K\] and the molar mass of nitrogen is, \[M = 28\]
Substitute the value of temperature and molar mass in above equation we get,
\[{\left( {{V_{rms}}} \right)_{{H_2}}} = \sqrt {\dfrac{{3R\left( {573} \right)}}{{28}}} \]
Now, the rms speed of hydrogen is,
\[{\left( {{V_{rms}}} \right)_{{H_2}}} = \sqrt {\dfrac{{3RT}}{M}} \]
The molar mass of hydrogen is, \[M = 2\]. When the rms speed of a \[{H_2}\] molecule is equal to the rms speed of a nitrogen molecule, then,
\[{\left( {{V_{rms}}} \right)_{{H_2}}} = {\left( {{V_{rms}}} \right)_{{N_2}}}\]
\[\Rightarrow \sqrt {\dfrac{{3RT}}{2}} = \sqrt {\dfrac{{3R\left( {573} \right)}}{{28}}} \]
\[\Rightarrow \dfrac{T}{2} = \dfrac{{573}}{{28}}\]
\[\Rightarrow T = \dfrac{{573 \times 2}}{{28}}\]
\[\therefore T = 40.92K\]
Therefore, the temperature at which the rms speed of a \[{H_2}\] molecule would be equal to the rms speed of a nitrogen molecule is 40.92K.
Note: The variable R is also used to denote the resistance. But in this case the R we are using is the universal gas constant and M is the molar mass. This universal gas constant value depends on the type of gas and remains the same every time.
Formula Used:
The formula for the root mean square speed is,
\[{V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \]
Where, R is universal gas constant, T is temperature and M is molar mass of gas.
Complete step by step solution:
Consider a nitrogen gas at \[{300^0}C\]. We need to find the value of temperature (in K) at which the rms speed of an \[{H_2}\] molecule would be equal to the RMS speed of a nitrogen molecule. For that, we have the formula for finding rms speed of nitrogen as,
\[{\left( {{V_{rms}}} \right)_{{N_2}}} = \sqrt {\dfrac{{3RT}}{M}} \]
We need to convert the temperature into kelvin, that is
\[T = \left( {300 + 273} \right)K\]
\[ \Rightarrow T = 573K\] and the molar mass of nitrogen is, \[M = 28\]
Substitute the value of temperature and molar mass in above equation we get,
\[{\left( {{V_{rms}}} \right)_{{H_2}}} = \sqrt {\dfrac{{3R\left( {573} \right)}}{{28}}} \]
Now, the rms speed of hydrogen is,
\[{\left( {{V_{rms}}} \right)_{{H_2}}} = \sqrt {\dfrac{{3RT}}{M}} \]
The molar mass of hydrogen is, \[M = 2\]. When the rms speed of a \[{H_2}\] molecule is equal to the rms speed of a nitrogen molecule, then,
\[{\left( {{V_{rms}}} \right)_{{H_2}}} = {\left( {{V_{rms}}} \right)_{{N_2}}}\]
\[\Rightarrow \sqrt {\dfrac{{3RT}}{2}} = \sqrt {\dfrac{{3R\left( {573} \right)}}{{28}}} \]
\[\Rightarrow \dfrac{T}{2} = \dfrac{{573}}{{28}}\]
\[\Rightarrow T = \dfrac{{573 \times 2}}{{28}}\]
\[\therefore T = 40.92K\]
Therefore, the temperature at which the rms speed of a \[{H_2}\] molecule would be equal to the rms speed of a nitrogen molecule is 40.92K.
Note: The variable R is also used to denote the resistance. But in this case the R we are using is the universal gas constant and M is the molar mass. This universal gas constant value depends on the type of gas and remains the same every time.
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