© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Write down the expression for RMS speed
The root-mean-square (rms) speed v_{\text{rms}} of a gas molecule of molar mass M (in kg/mol) at temperature T (in K) is given by:
v_{\text{rms}} = \sqrt{\frac{3 R T}{M}}
Here, R is the universal gas constant.
Step 2: Express RMS speed for N₂
Given that the temperature of nitrogen gas is 300^\circ \text{C} . First, convert this to Kelvin:
T_{\text{N}_2} = 300 + 273 = 573 \text{ K}
For nitrogen gas, whose molar mass is 28 \text{ g/mol} (i.e., 0.028 \text{ kg/mol} if we use SI units consistently), the rms speed is
v_{\text{N}_2} = \sqrt{\frac{3 R \times 573}{28}} \quad(\text{using the given mass in grams/mol in the fraction, consistent with the formula as shown})
Step 3: Express RMS speed for H₂
For hydrogen gas, the molar mass is 2 \text{ g/mol} . The temperature we need to find is T . So its rms speed is
v_{\text{H}_2} = \sqrt{\frac{3 R T}{2}}
Step 4: Set the two RMS speeds equal
We need v_{\text{N}_2} to be equal to v_{\text{H}_2} . Therefore,
\sqrt{\frac{3 R T}{2}} = \sqrt{\frac{3 R \times 573}{28}}
Squaring both sides eliminates the square root:
\frac{3 R T}{2} = \frac{3 R \times 573}{28}
Step 5: Solve for T
Cancel the common factors 3 R on both sides:
\frac{T}{2} = \frac{573}{28}
Thus,
T = 2 \times \frac{573}{28} = \frac{1146}{28}
Perform the division:
T \approx 41 \text{ K}
Step 6: Final Answer
Therefore, the required temperature at which the rms speed of a hydrogen molecule equals that of a nitrogen molecule at 300^\circ \text{C} is
T = 41 \text{ K}
