Answer:
\[P=\,\frac{1}{2}\,\rho
{{c}^{2}}\,=\,\frac{1}{2}\,\frac{M}{V}\,{{c}^{2}}\]
or \[c=\,\sqrt{\frac{2PV}{M}}\]
\[\therefore
\]\[\sqrt{\frac{{{c}_{2}}}{{{c}_{1}}}}=\sqrt{\frac{{{P}_{x}}{{V}_{2}}}{{{P}_{1}}{{V}_{1}}}}\] ?. (1)
Now
\[{{P}_{1}}{{V}_{1}}=R{{T}_{1}}\] and \[{{P}_{2}}{{V}_{2}}=R{{T}_{2}}\]
\[\therefore
\] \[\frac{{{P}_{2}}V{{ &
}_{2}}}{{{P}_{1}}{{V}_{1}}}\,=\,\frac{{{T}_{2}}}{{{T}_{1}}}\,=\,\frac{400}{300}\,=\,\frac{4}{3}\]
Hence
\[\frac{{{c}_{2}}}{{{c}_{1}}}\,=\frac{2}{\sqrt{3}}\] or \[{{c}_{2}}\,-\frac{2{{c}_{1}}}{\sqrt{3}}\,=\frac{200}{\sqrt{3}}\,m\,{{s}^{-1}}\]
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