Question
$4.0g$ of helium occupies $22400cm^3$ at STP. The specific heat capacity of helium at constant pressure is $5.0cal-K^{-1}mol^{-1}$. Calculate the speed of sound in helium at STP.
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Answer
$\mu=4\text{g}=4\times10^{-3}\text{kg},V = 22400cm^3 = 22400 \times 10^{-6}m^3$
$C_P = 5cal/mol-ki = 5 \times 4.2J/mol-k = 21J/mol-k$
$\text{C}_\text{P}=\frac{\gamma\text{R}}{\gamma-1}=\frac{\gamma\times8.3}{\gamma-1}$
$\Rightarrow21(\gamma-1)=\gamma(8.30$
$\Rightarrow21\gamma-21=8.3\gamma$
$\Rightarrow\gamma=\frac{21}{12.7}$
Since the condition is STP, $P = 1atm = 10^5pa$
$\text{V}=\sqrt{\frac{\gamma\text{f}}{\text{f}}}$
$=\sqrt{\frac{\frac{21}{12.7}\times10^5}{\frac{4\times10^{-3}}{22400\times10^{-6}}}}$
$=\sqrt{\frac{21\times10^5\times22400\times10^{-6}}{12.7\times4\times10^{-3}}}=962.28\text{m/s}$
$C_P = 5cal/mol-ki = 5 \times 4.2J/mol-k = 21J/mol-k$
$\text{C}_\text{P}=\frac{\gamma\text{R}}{\gamma-1}=\frac{\gamma\times8.3}{\gamma-1}$
$\Rightarrow21(\gamma-1)=\gamma(8.30$
$\Rightarrow21\gamma-21=8.3\gamma$
$\Rightarrow\gamma=\frac{21}{12.7}$
Since the condition is STP, $P = 1atm = 10^5pa$
$\text{V}=\sqrt{\frac{\gamma\text{f}}{\text{f}}}$
$=\sqrt{\frac{\frac{21}{12.7}\times10^5}{\frac{4\times10^{-3}}{22400\times10^{-6}}}}$
$=\sqrt{\frac{21\times10^5\times22400\times10^{-6}}{12.7\times4\times10^{-3}}}=962.28\text{m/s}$
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