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. 

Answer 2

The effective (root mean square) speed of a gas molecule follows \( v \propto \sqrt{\dfrac{T}{M}} \), where \(T\) is the absolute temperature and \(M\) is the molar mass. Hydrogen has \(M_H = 2\) and oxygen has \(M_O = 32\), and oxygen is at \(T_O = 47^{\circ}\text{C} = 320\,\text{K}\). Because the two effective speeds are equal, \( \sqrt{\dfrac{T_H}{2}} = \sqrt{\dfrac{320}{32}} \), which fixes the required hydrogen temperature. Let's check each option against this condition.

1. 60 K: Putting \(T_H = 60\) gives \( \sqrt{60/2} = \sqrt{30} \approx 5.48 \), while the oxygen side gives \( \sqrt{320/32} = \sqrt{10} \approx 3.16 \). These are not equal, so 60 K does not satisfy the condition.
2. 40 K: Putting \(T_H = 40\) gives \( \sqrt{40/2} = \sqrt{20} \approx 4.47 \), still not equal to \( \sqrt{10} \approx 3.16 \), so 40 K also fails the equality even though it looks close to the working.
3. 20 K: Putting \(T_H = 20\) gives \( \sqrt{20/2} = \sqrt{10} \approx 3.16 \), which exactly matches \( \sqrt{320/32} = \sqrt{10} \). This value satisfies the equal-speed condition.
4. 0 K: At \(T_H = 0\), the hydrogen speed would be zero, which cannot equal a finite oxygen speed. So this option is ruled out immediately.

Only \( T_H = 20\,\text{K} \) makes \( \sqrt{T_H/M_H} \) equal to \( \sqrt{T_O/M_O} \), so it is the temperature at which hydrogen's effective speed matches oxygen's at 320 K.

So the correct answer is 20 K.