The temperature of a gas is $-50^\circ ext{C}$. To what temperature the gas should be heated so that the RMS

Question

The temperature of a gas is $-50^\circ	ext{C}$. To what temperature the gas should be heated so that the RMS speed is increased by 3 times?

Accepted Answer

The root mean square speed ($v_{rms}$) is directly proportional to the square root of the absolute temperature ($v_{rms} \propto \sqrt{T}$).

1.  **Initial State:**
    $T_1 = -50^\circ	ext{C} = 273 - 50 = 223 	ext{ K}$.
    Let the initial speed be $v_1 = v$.

2.  **Final State:**
    The speed is increased *by* 3 times. This means the final speed $v_2 = v + 3v = 4v$.

3.  **Calculation:**
    Using the relation $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$:
    $\frac{4v}{v} = \sqrt{\frac{T_2}{223}}$
    $4 = \sqrt{\frac{T_2}{223}}$
    Squaring both sides:
    $16 = \frac{T_2}{223}$
    $T_2 = 16 	imes 223 = 3568 	ext{ K}$

4.  **Conversion:**
    Convert the final temperature back to Celsius:
    $T_2 (	ext{in } ^\circ	ext{C}) = 3568 - 273 = 3295^\circ	ext{C}$.

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