Velocity Dependence on Temperature

The velocity of sound ($v$) in an ideal gas is directly proportional to the square root of the absolute temperature ($T$) in Kelvin. The formula is:

$v \propto \sqrt{T}$

This can be written as $v = k\sqrt{T}$, where $k$ is a constant for a given gas.

Temperature Conversion

Temperatures must be in Kelvin for the formula. The conversion from Celsius ($^\circ\text{C}$) to Kelvin (K) is:

$T(\text{K}) = T(^\circ\text{C}) + 273.15$

Calculation for $\alpha$

Let $v_0$ be the velocity at $0^\circ\text{C}$ and $v_\alpha$ be the velocity at $\alpha^\circ\text{C}$.

The corresponding absolute temperatures are:

• $T_0 = 0^\circ\text{C} = 0 + 273.15 = 273.15 \text{ K}$
• $T_\alpha = \alpha^\circ\text{C} = (\alpha + 273.15) \text{ K}$

According to the problem, the velocity doubles:

$v_\alpha = 2 v_0$

Using the proportionality $v \propto \sqrt{T}$:

$\frac{v_\alpha}{v_0} = \frac{k\sqrt{T_\alpha}}{k\sqrt{T_0}} = \sqrt{\frac{T_\alpha}{T_0}}$

Since $\frac{v_\alpha}{v_0} = 2$:

$2 = \sqrt{\frac{T_\alpha}{T_0}}$

Squaring both sides:

$4 = \frac{T_\alpha}{T_0}$

$T_\alpha = 4 T_0$

Substitute the temperatures in Kelvin:

$\alpha + 273.15 = 4 \times 273.15$

Solve for $\alpha$:

$\alpha = (4 \times 273.15) - 273.15$

$\alpha = 3 \times 273.15$

$\alpha = 819.45$

The value of $\alpha$ is approximately 819.45. Rounding to the nearest integer gives $\alpha = 819$.