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 $\alpha = 819$.
Similar Questions
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Important Questions from Oscillations and Waves
- Two tuning forks $A$ and $B$ are sounded together giving rise to $8$ beats in $2\text{ s}$. When fork $A$ is loaded with wax, the beat frequency is reduced to $4$ beats in $2\text{ s}$. If the original frequency of tuning fork $B$ is $380\text{ Hz}$ then original frequency of tuning fork $A$ is _________ $\text{Hz}$.
- 5 beats/second are heard when a turning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either 0.95 m or 1 m. The frequency of the fork will be
- A sinusoidal wave of wavelength 7.5 cm travels a distance of 1.2 cm along the x-direction in 0.3 sec. The crest P is at x=0 at t=0 sec and maximum displacement of the wave is 2 cm. Which equation correctly represents this wave ?
A particle is subjected to two simple harmonic motions as :
$x_1 = \sqrt{7} \sin 5t \text{ cm}$ and $x_2 = 2\sqrt{7} \sin \left(5t + \frac{\pi}{3}\right) \text{ cm}$
where x is displacement and t is time in seconds. The maximum acceleration of the particle is $x \times 10^{-2} \text{ ms}^{-2}$. The value of x is :- The equation of a transverse wave is $y = y_0 \sin 2\pi(ft - \frac{x}{\lambda})$. If the maximum particle velocity be four times that of wave velocity then
