Question

To what temperature should the hydrogen at $ 327^{\circ}C $ be cooled at constant pressure, so that the root mean square velocity of its molecules becomes half of its previous value ?

• A. $ -123^{\circ} C $
• B. $ 123^{\circ} C $
• C. $ -100^{\circ} C $
• D. $ 0^{\circ} C $

Answer 1

The Correct Option is A

$v_{ rms } \propto \sqrt{\frac{3 R T}{M}}$ $\Rightarrow T \propto v_{ rms }^{2}$ $\Rightarrow \frac{T_{2}}{T_{1}}=\left[\frac{v_{2}}{v_{1}}\right]^{2}$ $=\frac{1}{4}$ $\Rightarrow T_{2}=\frac{T_{1}}{4}=\frac{(273+327)}{4}$ $=150 K =-123^{\circ} C$

Answer 2

Given:
Initial temperature, \(T_1 = 327^\circ \text{C}\)
\(T_1 = 327 + 273.15 = 600.15 \text{ K}\)

Relationship with rms velocity:
The root mean square (rms) velocity \(v_{rms}\) of gas molecules is related to temperature T by:
\(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\)

Given condition and the relation:
\(\frac{v_{\text{rms}2}}{v_{\text{rms}1}} = \sqrt{\frac{T_2}{T_1}}\)

Given \(v_{\text{rms}2} = \frac{1}{2} v_{\text{rms}1}\):
\(\frac{1}{2} = \sqrt{\frac{T_2}{T_1}}\)

Square both sides to solve for \(\frac{T_2}{T_1}\):
\(\left( \frac{1}{2} \right)^2 = \frac{T_2}{T_1}\)
\(\frac{1}{4} = \frac{T_2}{T_1}\)

Solve for \(T_2\):
\(T_2 = \frac{1}{4} T_1\)

\(T_2 = \frac{1}{4} \times 600.15\)
\(T_2 = 150.0375 \text{ K}\)
\(T_2 = 150.0375 - 273.15\)
\(T_2 \approx -123.1125^\circ \text{C}\)

So, the correct option is (A): \(-123^\circ \text{C}\)