At which temperature the r.m.s. velocity of a hydrogen molecule equal to that of an oxygen molecule at 47ºC?
- 80 K
- -73 K
- 20 K
- 4 K
The Correct Option is C
Approach Solution - 1
Using the Formula for Root Mean Square (r.m.s.) Velocity: The r.m.s. velocity \( v_{rms} \) for a gas is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas.
Set up the Equation for Hydrogen and Oxygen: To find the temperature at which the r.m.s. velocity of hydrogen equals that of oxygen at 47°C, we set:
\[ \sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}} \]
Isolate \( T_{H_2} \): Square both sides to remove the square root:
\[ \frac{3RT_{H_2}}{M_{H_2}} = \frac{3RT_{O_2}}{M_{O_2}} \]
Simplify by canceling \( 3R \) on both sides:
\[ T_{H_2} = T_{O_2} \times \frac{M_{H_2}}{M_{O_2}} \]
Substitute Values for Molar Mass and Temperature: Given \( T_{O_2} = 47°C = 320 \, K \),
\[ T_{H_2} = 320 \times \frac{2}{32} = 20 \, K \]
Approach Solution -2
To find the temperature at which the root mean square (r.m.s.) velocity of a hydrogen molecule equals that of an oxygen molecule at 47°C, we use the formula for r.m.s. velocity:
\(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\)
where:
- \(v_{\text{rms}}\) is the root mean square velocity,
- \(k\) is the Boltzmann constant,
- \(T\) is the temperature in Kelvin,
- \(m\) is the molecular mass.
Given that the r.m.s velocity of the hydrogen molecule is equal to that of the oxygen molecule, we can set up the equation:
\(\sqrt{\frac{3kT_{\text{H}_2}}{m_{\text{H}_2}}} = \sqrt{\frac{3kT_{\text{O}_2}}{m_{\text{O}_2}}}\)
From this equation, we cancel out the common terms:
\(\frac{T_{\text{H}_2}}{m_{\text{H}_2}} = \frac{T_{\text{O}_2}}{m_{\text{O}_2}}\)
Rearranging gives us:
\(T_{\text{H}_2} = T_{\text{O}_2} \times \frac{m_{\text{H}_2}}{m_{\text{O}_2}}\)
The molecular masses are:
- Molecular mass of hydrogen, \(m_{\text{H}_2} = 2 \text{ u}\).
- Molecular mass of oxygen, \(m_{\text{O}_2} = 32 \text{ u}\).
The temperature of the oxygen molecule, \(T_{\text{O}_2}\), must be in Kelvin. 47°C is converted to Kelvin as follows:
\(T_{\text{O}_2} = 47 + 273 = 320 \text{ K}\)
Substituting these values into the equation:
\(T_{\text{H}_2} = 320 \times \frac{2}{32}\)
Simplifying gives:
\(T_{\text{H}_2} = 320 \times \frac{1}{16} = 20 \text{ K}\)
Therefore, the correct answer is 20 K.
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Concepts Used:
Gas Laws
The gas laws were developed at the end of the 18th century, when scientists began to realize that relationships between pressure, volume and temperature of a sample of gas could be obtained which would hold to approximation for all gases.
The five gas laws are:
- Boyle’s Law, which provides a relationship between the pressure and the volume of a gas.
- Charles’s Law, which provides a relationship between the volume occupied by a gas and the absolute temperature.
- Gay-Lussac’s Law, which provides a relationship between the pressure exerted by a gas on the walls of its container and the absolute temperature associated with the gas.
- Avogadro’s Law, which provides a relationship between the volume occupied by a gas and the amount of gaseous substance.
- The Combined Gas Law (or the Ideal Gas Law), which can be obtained by combining the four laws listed above.