The molecules of a given mass of a gas have r.m.s velocity of 200 ms−1 at 27 ∘C and pressure of 1.0×105 Nm−2. When the temperature and pressure of the gas are respectively 127 ∘C and 0.05×105 Nm−2 ,the r.m.s velocity in ms−1.

Have v1,rms=200 m/stemperature T1=27∘C=27+273=300 Ktemperature T2=127∘C=127+273=400 K, v2,rms=?R.M.S. Velocity, v∝T⇒v2,rms200=400300⇒v2,rms=200×23 m/s⇒v=4003 m/s

The molecules of a given mass of a gas have r.m.s velocity of 200 ms−1 at 27 ∘C and pressure of 1.0×105 Nm−2. When the temperature and pressure of the gas are respectively 127 ∘C and 0.05×105 Nm−2 ,the r.m.s velocity in ms−1.

Have v1,rms=200 m/stemperature T1=27∘C=27+273=300 Ktemperature T2=127∘C=127+273=400 K, v2,rms=?R.M.S. Velocity, v∝T⇒v2,rms200=400300⇒v2,rms=200×23 m/s⇒v=4003 m/s

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The molecules of a given mass of a gas have $r . m . s$ velocity of $200 {ms}^{- 1} at 27^{\circ} C and pressure of 1.0 \times 10^{5} {Nm}^{- 2} .$ When the temperature and pressure of the gas are respectively $127^{\circ} C and 0.05 \times 10^{5} {Nm}^{- 2}$ ,the $r . m . s$ velocity in ${ms}^{- 1}$ .

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$100 \sqrt{2}$

$\frac{400}{\sqrt{3}}$

$\frac{100 \sqrt{2}}{3}$

$\frac{100}{3}$

answer is B.

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Detailed Solution

Have $v_{1, r m s} = 200 m / s$

temperature $T_{1} = 27^{\circ} C = 27 + 273 = 300 K$

temperature $T_{2} = 127^{\circ} C = 127 + 273 = 400 K, v_{2, rms} = ?$

R.M.S. Velocity, $v \propto \sqrt{T}$

$\begin{array}{l} \Rightarrow \frac{v_{2, rms}}{200} = \sqrt{\frac{400}{300}} \\ \Rightarrow v_{2, rms} = \frac{200 \times 2}{\sqrt{3}} m / s \Rightarrow v = \frac{400}{\sqrt{3}} m / s \end{array}$