To determine the speed of sound in oxygen at Standard Temperature and Pressure (S.T.P.), we need to use the formula for the speed of sound in gases:
\(v = \sqrt{\frac{\gamma \cdot R \cdot T}{M}}\)
where:
For oxygen, the values are:
At S.T.P., the temperature \(T\) is generally taken as \(273 \, \text{K}\).
Substituting these values into the formula:
\(v = \sqrt{\frac{1.4 \cdot 8.3 \cdot 273}{0.032}}\)
Calculating inside the square root:
\(v = \sqrt{\frac{3170.44}{0.032}}\)
\(v = \sqrt{99076.25}\)
\(v \approx 314.8 \, \text{m/s}\)
Given the options, the closest value is \(310 \, \text{m/s}\), which can be considered as the correct approximation under the given conditions and assumptions.
Therefore, the correct answer is:
\(310 \, \text{m/s}\)
The speed of sound v in a gas is given by:
\[ v = \sqrt{\frac{\gamma RT}{M}}, \]
where:
We are given:
\[ \gamma = 1.4, \quad R = 8.3 \, \text{J/K mol}, \quad T = 273 \, \text{K}, \quad M = 32 \times 10^{-3} \, \text{kg/mol}. \]
Substitute these values into the formula:
\[ v = \sqrt{\frac{1.4 \times 8.3 \times 273}{32 \times 10^{-3}}}. \]
Calculate the expression inside the square root:
\[ 1.4 \times 8.3 = 11.62, \]
\[ 11.62 \times 273 = 3173.26, \]
\[ \frac{3173.26}{32 \times 10^{-3}} = 99164.375. \]
Now take the square root to find v:
\[ v = \sqrt{99164.375} \approx 315 \, \text{m/s}. \]
The closest answer to this calculated value is: 310 m/s.

Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 