Question
Use the formula $\text{v}=\sqrt{\frac{\gamma\text{P}}{\rho}}$ to explain why the speed of sound in air:
Increases with temperature,
Increases with temperature,
✓
Answer
Take the relation:
$\text{v}=\sqrt{\frac{\gamma\text{P}}{\rho}}\ \dots(\text{i})$
or one mole of any ideal gas, the equation can be written as:
PV = RT
$\text{P}=\frac{\text{RT}}{\text{V}}\ \dots(\text{ii})$
Substituting equation (ii) in equation (i), we get:
$\text{v}=\sqrt{\frac{\gamma\text{RT}}{\text{VP}}}=\sqrt{\frac{\gamma\text{RT}}{\text{M}}}\ \dots(\text{iii})$
where,
mass, $\text{M}=\rho\text{V}$ is a constant
γ and R are also constants
We conclude from equation (iii) that $\text{v}\propto\sqrt{\text{T}}$
Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium, i.e., the speed of the sound increases with an increase in the temperature of the gaseous medium and vice versa.
$\text{v}=\sqrt{\frac{\gamma\text{P}}{\rho}}\ \dots(\text{i})$
or one mole of any ideal gas, the equation can be written as:
PV = RT
$\text{P}=\frac{\text{RT}}{\text{V}}\ \dots(\text{ii})$
Substituting equation (ii) in equation (i), we get:
$\text{v}=\sqrt{\frac{\gamma\text{RT}}{\text{VP}}}=\sqrt{\frac{\gamma\text{RT}}{\text{M}}}\ \dots(\text{iii})$
where,
mass, $\text{M}=\rho\text{V}$ is a constant
γ and R are also constants
We conclude from equation (iii) that $\text{v}\propto\sqrt{\text{T}}$
Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium, i.e., the speed of the sound increases with an increase in the temperature of the gaseous medium and vice versa.
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600
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800
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1000
|
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|
10
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20
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40
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20
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10
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