Question

At temperature 'T', the 'effective' speed of gaseous hydrogen molecules (molecular weight = 2) is equal to that of oxygen molecules (molecular weight = 32) at 47°C. The value of 'T' is:

• A. 60 K
• B. 40 K
• C. 20 K
• D. 0 K

Hint:

The effective speed of gas molecules is inversely proportional to the square root of the molecular weight.

Answer 1

The Correct Option is B

The speed of gas molecules is related to the temperature and molecular weight by the formula: \[ v \propto \sqrt{\frac{T}{M}} \] where \( v \) is the speed, \( T \) is the temperature, and \( M \) is the molecular weight. Since the speeds of hydrogen and oxygen molecules are equal, we can write: \[ \frac{v_H}{v_O} = \sqrt{\frac{T_H}{M_H}} \div \sqrt{\frac{T_O}{M_O}} \] Simplifying for the given molecular weights and temperatures, we find: \[ \frac{v_H}{v_O} = \sqrt{\frac{T_H}{T_O}} \times \sqrt{\frac{M_O}{M_H}} \] Substituting the given values: \[ \sqrt{\frac{T_H}{T_O}} \times \sqrt{\frac{32}{2}} = 1 \] Solving for \( T_H \): \[ T_H = 40 \, \text{K} \] Thus, the temperature \( T \) is 40 K.