Question 8 of 9intermediate🔧 ApplyNumerical3 marks

At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at –20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).

Correct Answer

The temperature of argon gas is 2524 K (approximately).

Given:

  • Temperature of helium gas, T(He) = –20 °C = 253 K
  • Atomic mass of Ar = 39.9 u
  • Atomic mass of He = 4.0 u

Step 1: Formula for rms speed The root mean square speed is given by:

vrms=3kBTmv_{rms} = \sqrt{\frac{3k_BT}{m}}

Step 2: Condition for equal rms speeds For (vrms)Ar=(vrms)He(v_{rms})_{Ar} = (v_{rms})_{He}:

3kBTArmAr=3kBTHemHe\sqrt{\frac{3k_BT_{Ar}}{m_{Ar}}} = \sqrt{\frac{3k_BT_{He}}{m_{He}}}

Step 3: Solving for T(Ar) Squaring both sides and simplifying:

TArMAr=THeMHe\frac{T_{Ar}}{M_{Ar}} = \frac{T_{He}}{M_{He}}

TAr=THe×MArMHeT_{Ar} = T_{He} \times \frac{M_{Ar}}{M_{He}}

TAr=253×39.94.0T_{Ar} = 253 \times \frac{39.9}{4.0}

Step 4: Final calculation TAr=253×9.975=2523.68T_{Ar} = 253 \times 9.975 = 2523.68 K ≈\approx 2524 K

Exercise: EXERCISES | Q: 12.9
For More Understanding

Explanation

This question tests the understanding of rms speed formula and its dependence on temperature and molecular mass. From the context, the formula 12mvrms2=32kBT\frac{1}{2} m v_{rms}^2 = \frac{3}{2} k_B T establishes that rms speed depends on both temperature and mass. For equal rms speeds, the ratio T/M must be equal for both gases. Since argon is approximately 10 times heavier than helium, it needs to be at roughly 10 times the temperature to have the same rms speed.

Solution Steps

  1. Step 1: Convert temperature of He from –20 °C to Kelvin: T(He) = 253 K

  2. Step 2: Write the condition for equal rms speeds: 3kBTArmAr=3kBTHemHe\sqrt{\frac{3k_BT_{Ar}}{m_{Ar}}} = \sqrt{\frac{3k_BT_{He}}{m_{He}}}

  3. Step 3: Simplify to get: TArMAr=THeMHe\frac{T_{Ar}}{M_{Ar}} = \frac{T_{He}}{M_{He}}

  4. Step 4: Calculate T(Ar) = 253 × (39.9/4.0) = 2524 K

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