Regarding the speed of sound in a gas, match | vedclass

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Question

(A) The speed of sound in an ideal gas is given by $v = \sqrt{\frac{\gamma RT}{M_W}}$, where $\gamma$ is the adiabatic index, $R$ is the gas constant,

Vedclass · Accepted Answer

$A-Q, B-R, C-Q, D-S$

Vedclass · Answer

(A) The speed of sound in an ideal gas is given by $v = \sqrt{\frac{\gamma RT}{M_W}}$, where $\gamma$ is the adiabatic index, $R$ is the gas constant, $T$ is the absolute temperature, and $M_W$ is the molar mass.$A$. If $T$ becomes $4T$, then $v' = \sqrt{\frac{\gamma R(4T)}{M_W}} = 2v$. Since pressure $P$ does not affect the speed of sound in an ideal gas (as $P \propto \rho T$ at constant $M_W$), the speed becomes $2$ times. Thus, $A-Q$.$B$. If only pressure is changed, $T$ remains constant. Since $v \propto \sqrt{T}$, the speed remains unchanged. Thus, $B-R$.$C$. If $T$ becomes $4T$, $v' = \sqrt{\frac{\gamma R(4T)}{M_W}} = 2v$. The speed becomes $2$ times. Thus, $C-Q$.$D$. If $M_W$ becomes $4M_W$, then $v' = \sqrt{\frac{\gamma RT}{4M_W}} = \frac{1}{2}v$. The speed becomes half. Thus, $D-S$.Therefore, the correct matching is $A-Q, B-R, C-Q, D-S$.

$A$. Temperature of gas is made $4$ times and pressure $2$ times	$P$. Speed becomes $2\sqrt{2}$ times
$B$. Only pressure is made $4$ times without change in temperature	$Q$. Speed becomes $2$ times
$C$. Only temperature is changed to $4$ times	$R$. Speed remains unchanged
$D$. Molecular mass of the gas is made $4$ times	$S$. Speed becomes half

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