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.
For an ideal gas, a cyclic process ABCA as shown in the P–T diagram. When represented in P–V plot, it would be 

The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9 V with internal resistance of 1 \(\Omega\) is connected across these points is ______ J. 
The given circuit works as: 
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}