If the kinetic energy of \(\mathrm{O}_2\) gas is \(4.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\), its RMS speed in…
If the kinetic energy of \(\mathrm{O}_2\) gas is \(4.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\), its RMS speed in \(\mathrm{cm} \mathrm{s}^{-1}\) is
- \(5.0 \times 10^2\)
- \(5.0 \times 10^3\)
- \(5.0 \times 10^4\)
- \(5.0 \times 10^4\)
Solution
Kinetic energy of \(\mathrm{O}_2=4.0 \mathrm{~kJ} / \mathrm{mol}\)
\(\begin{aligned}
\mathrm{KE} & =\frac{3 R T}{2} \text { and } v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}} \\
v_{\mathrm{rms}} & =\sqrt{\frac{2 \mathrm{KE}}{M}} \\
& =\sqrt{\frac{2 \times 4 \times 10^3 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2} \mathrm{~mol}^{-1}}{32 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}^{-1}}}
\end{aligned}\)
\(\begin{aligned} & {\left[\because 1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}\right] } \\ = & \frac{10^3}{2}=5 \times 10^2 \mathrm{~m} / \mathrm{s}\end{aligned}\)
\(v_{\text {rms }}\) in \(\mathrm{cm} / \mathrm{s}=50 \times 10^4 \mathrm{~cm} / \mathrm{s}\)
Asked in: AP EAMCET 2019 (20 Apr Shift 1)
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