Question

An ideal gas with density (3.0 g L^-1) has a pressure of (684 mm Hg) at (25^C). The rms speed (in m s^-1) of the gas is (1 atm = 10^5 Pa).

• A. \(3 \times 10^{2}\)
• B. \(3 \times 10^{3}\)
• C. \(4 \times 10^{2}\)
• D. \(4 \times 10^{3}\)

Hint:

For rms speed using density, directly use: \[ u_{\text{rms}} = \sqrt{\frac{3P}{\rho}} \] This avoids unnecessary use of \(R\) and \(M\) and is the fastest exam method.

Answer 1

The Correct Option is A

Concept: The root mean square (rms) speed of an ideal gas is given by: \[ u_{\text{rms}} = \sqrt{\frac{3RT}{M}} \] Using the ideal gas equation \(PV = nRT\), we write: \[ P = \frac{\rho RT}{M} \Rightarrow \frac{RT}{M} = \frac{P}{\rho} \] Substituting into the rms expression: \[ u_{\text{rms}} = \sqrt{\frac{3P}{\rho}} \]

Step 1: Conversion of pressure into SI units
Given: \[ P = 684 \,\text{mm Hg} \] Using \(760 \,\text{mm Hg} = 1 \,\text{atm} = 10^{5} \,\text{Pa}\): \[ P = \frac{684}{760} \times 10^{5} = 0.9 \times 10^{5} = 9 \times 10^{4} \,\text{Pa} \]

Step 2: Conversion of density into SI units
Given: \[ \rho = 3.0 \,\text{g L}^{-1} \] Convert: \[ 1 \,\text{g L}^{-1} = 1 \,\text{kg m}^{-3} \] So, \[ \rho = 3.0 \,\text{kg m}^{-3} \]

Step 3: Substituting values in rms formula
\[ u_{\text{rms}} = \sqrt{\frac{3P}{\rho}} = \sqrt{\frac{3 \times 9 \times 10^{4}}{3}} \] Cancel 3: \[ u_{\text{rms}} = \sqrt{9 \times 10^{4}} \] \[ u_{\text{rms}} = 3 \times 10^{2} \,\text{m s}^{-1} \]

Final Answer: \[ \boxed{3 \times 10^{2} \,\text{m s}^{-1}} \]