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A glass bulb of 1 litre capacity contains $\small 2\times10^{21}$ molecules of nitrogen exerting pressure of $\small 7.57\times 10^3Nm^{-2}$ . After calculating the root mean square speed (in m/sec) and the temperature of gas molecules finally calculate $n_{mp}$ for these molecules at this temperature, if the ratio of $\dpi{100} \small u_{mp}$ to $\small u_{rms}$ is 0.82.
Option: 1
40.52

Option: 2
503.56

Option: 3
405.26

Option: 4
400.43

Answers (1)

best_answer

We have given:

$P=7.57\times10^3 Nm^{-2}$

$V=10^{-3}m^3$

And, $n=(2\times 10^{21})/(6.023\times 10^{23})$

Now, we know the ideal gas equation,
PV = nRT

$\\\mathrm{7.57\: x\: 10^{3}\: x\: 10^{-3}\: =\: \frac{2\: x\: 10^{21}}{6.023\: x\: 10^{23}}\: x\: 8.314\: x\: T}$

Thus, T = 274.2K
Now, again we know that

$u_{rms}=\sqrt{\frac{3RT}{M}}$

$u_{rms}=\sqrt{\frac{3\times 8.314 \times 274.2}{28\times10^{-3}}}$

$u_{rms}=494.22m/sec$

Now, as we have given:
$u_{mp}=0.82u_{rms}$
Thus, $u_{mp}=0.82\times 494.22$
Hence, $u_{mp}=0.82\times 494.22=405.26m/sec$

Posted by

Rakesh

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