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

Step 1: Recall the rms speed formula.
The root mean square speed of gas particles is \[ u_{rms} = \sqrt{\frac{3RT}{M}} \] where $T$ is temperature and $M$ is molar mass. We can rewrite it in a handier form using density.

Step 2: Bring in density.
From the ideal gas law $PV=nRT$, we can show that $P = \dfrac{\rho RT}{M}$, so $\dfrac{RT}{M} = \dfrac{P}{\rho}$. Putting this inside the speed formula gives \[ u_{rms} = \sqrt{\frac{3P}{\rho}} \] This is neat because it needs only pressure and density.

Step 3: Convert the pressure to SI units.
Given pressure is $684$ mm Hg. Since $760$ mm Hg equals $10^5$ Pa, \[ P = \frac{684}{760} \times 10^5 = 0.9 \times 10^5 \text{ Pa} \]

Step 4: Note the density in SI units.
The density is $3.0$ g per litre, which equals $3.0$ kg per cubic metre. So $\rho = 3.0 \text{ kg/m}^3$.

Step 5: Substitute into the formula.
\[ u_{rms} = \sqrt{\frac{3 \times 0.9 \times 10^5}{3.0}} = \sqrt{\frac{2.7 \times 10^5}{3.0}} \] \[ = \sqrt{0.9 \times 10^5} = \sqrt{9 \times 10^4} \]

Step 6: Take the square root and finish.
\[ u_{rms} = 3 \times 10^2 \text{ m/s} \] So the rms speed of the gas is about three hundred metres per second.
\[ \boxed{3 \times 10^{2}\ \text{m s}^{-1}} \]