Question

If the rms speed of the molecules of a gas at a temperature of 77 $^\circ$C is 50 ms$^{-1}>$, then the rms speed of the same gas molecules at a temperature of 150.5 $^\circ$C is

• A. 65 ms$^{-1}$
• B. 35 ms$^{-1}$
• C. 55 ms$^{-1}$
• D. 45 ms$^{-1}$

Hint:

In kinetic theory of gases, always use absolute temperature (Kelvin). The rms speed $v_{rms}$ is proportional to $\sqrt{T}$. For ratio-based problems, this proportionality is all you need: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.

Answer 1

The Correct Option is C

The root mean square (rms) speed of gas molecules is given by the formula:
$v_{rms} = \sqrt{\frac{3RT}{M}}$, where T is the absolute temperature in Kelvin.
From this formula, we can see that the rms speed is proportional to the square root of the absolute temperature: $v_{rms} \propto \sqrt{T}$.
We are given two states of the gas. Let's denote them by subscripts 1 and 2.
State 1:
Temperature $T_1 = 77 ^\circ\text{C}$. We must convert this to Kelvin: $T_1 = 77 + 273 = 350$ K.
RMS speed $v_1 = 50$ m/s.
State 2:
Temperature $T_2 = 150.5 ^\circ\text{C}$. Convert to Kelvin: $T_2 = 150.5 + 273 = 423.5$ K.
RMS speed is $v_2$, which we need to find.
Using the proportionality, we can set up a ratio:
$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
$\frac{v_2}{50} = \sqrt{\frac{423.5}{350}}$.
$\frac{v_2}{50} = \sqrt{\frac{847}{700}} = \sqrt{\frac{121 \times 7}{100 \times 7}} = \sqrt{\frac{121}{100}} = \frac{11}{10}$.
Solving for $v_2$:
$v_2 = 50 \times \frac{11}{10} = 5 \times 11 = 55$ m/s.